Redox Reactions and Electrochemistry NEET 2007: Solution

Q: What is the answer to the NEET 2007 copper–silver equilibrium question?

Option D, 4.0 × 10^15, is correct. Using E° = +0.46 V and n = 2 gives log₁₀K ≈ 15.57 and K ≈ 3.7 × 10^15. Option D is the closest listed value, not the exact calculated result.

Q: Why is n equal to 2 in the copper–silver reaction?

One copper atom loses two electrons as it changes from Cu to Cu²⁺. Two Ag⁺ ions each accept one electron, so the balanced reaction transfers two electrons overall. Do not add electrons lost and gained: these are the same two electrons.

Q: How do I calculate the equilibrium constant from standard cell potential at 298 K?

Use E° = (0.0591/n) log₁₀K, with E° in volts and n taken from the balanced reaction. Rearrange to log₁₀K = nE°/0.0591, then take the base-10 antilog. For n = 2 and E° = +0.46 V, K ≈ 3.7 × 10^15.

Q: What happens to E° and K when a reaction is reversed or doubled?

Reversing a reaction changes the sign of E° and replaces K with 1/K, while the electron count remains unchanged. Doubling the balanced equation doubles the electron count and squares K, but leaves E° unchanged. For the doubled copper–silver reaction, n = 4 and E° remains +0.46 V.

What is the answer to the Redox Reactions and Electrochemistry NEET 2007 question?

Option D is correct for the copper–silver equilibrium-constant MCQ in Redox Reactions and Electrochemistry NEET 2007. Find the equilibrium constant at 298 K when copper metal reacts with aqueous silver ions, given a standard cell potential of +0.46 V:

C u (s) + 2 A g^{+} (a q) \to C u^{2 +} (a q) + 2 A g (s)

The options are:

A $2.4 \times 10^{10}$
B $2.0 \times 10^{10}$
C $4.0 \times 10^{10}$
D $4.0 \times 10^{15}$

This 2007 Chemistry question is tagged hard by this question bank, with a bank-set target time of 90 seconds. Neither is an official exam classification. The solution needs three decisions: count transferred electrons, calculate the logarithm, then match the complete scientific-notation value.

Why are two electrons used in the copper–silver reaction?

Copper changes from oxidation state zero to plus two, so each copper atom loses two electrons. This establishes the electron count for the balanced reaction as written, before numerical substitution.

The oxidation half-reaction is:

C u (s) \to C u^{2 +} (a q) + 2 e^{-}

Each silver ion accepts one electron. The supplied reaction contains two silver ions, so the balanced reduction half-reaction is:

2 A g^{+} (a q) + 2 e^{-} \to 2 A g (s)

Adding the half-reactions cancels two electrons and recovers the supplied overall reaction:

C u (s) + 2 A g^{+} (a q) \to C u^{2 +} (a q) + 2 A g (s)

Therefore: $n = 2$

Count the electrons transferred, not their appearances on the page. The electron count is neither the charge on one silver ion nor four from adding electrons lost and gained. Both half-reactions describe the same two electrons moving from copper to silver ions.

How do you calculate the equilibrium constant’s logarithm at 298 K?

The base-10 logarithm of the equilibrium constant is approximately 15.57. Use the standard overall cell potential, not an isolated half-cell potential, and retain the official solution’s 0.0591 factor.

At 298 K, with the standard potential expressed in volts:

E^{\circ} = \frac{0.0591}{n} \log_{10} K

Rearrange before inserting numbers:

\log_{10} K = \frac{n E^{\circ}}{0.0591}

Substitute the two-electron count and supplied potential:

\log_{10} K = \frac{2 \times 0.46}{0.0591} = \frac{0.92}{0.0591} \approx 15.57

Here, “log” means base 10, not the natural logarithm. Do not replace 0.0591 with another rounded factor midway through the calculation.

A useful sign check is:

E^{\circ} > 0 \Rightarrow K > 1

This cannot select an option here. Every listed value exceeds one, so the magnitude still has to be calculated.

How does the antilog give option D?

D is the closest listed value, not the exact calculated value. To take the antilog, split the logarithm into its integer and fractional parts:

K = 10^{15.57} = 10^{15} \times 10^{0.57} \approx 3.7 \times 10^{15}

The integer part fixes the power of ten; the fractional part supplies the coefficient. The official answer is therefore:

K \approx 3.7 \times 10^{15} \Rightarrow D: 4.0 \times 10^{15}

These are not conflicting answers. One is the worked approximation; the other is the closest option.

After the full calculation, use this calculator-free check: $0.0591 \times 15 = 0.8865$ $0.0591 \times 16 = 0.9456$

Since the numerator is 0.92: $15 < \log_{10} K < 16$

Only D lies in the resulting range. All three options with a tenth-power exponent fail this check.

Why can matching only the coefficient lead to option C?

A solver could hypothetically obtain a coefficient near 3.7, round it towards 4.0, then select C without checking the exponent. That is incomplete reconstruction of the antilog, followed by coefficient-only option matching. $10^{0.57} \approx 3.7$

This is only the fractional-part contribution, not the complete equilibrium constant. Compare the two options:

C = 4.0 \times 10^{10}, D = 4.0 \times 10^{15}

\frac{D}{C} = 10^{5} = 100, 000

D is 100,000 times C despite their identical coefficients. Checking C directly exposes the error:

\log_{10} (4.0 \times 10^{10}) = 10 + \log_{10} 4 \approx 10.60 \neq 15.57

Write the complete scientific-notation result before scanning the options. Matching the coefficient is not enough: the exponent must match too.

What happens if you reverse the reaction or double it?

Reversing the reaction changes the sign of the standard potential and takes the reciprocal of the equilibrium constant. Doubling the equation doubles the electron count and squares the equilibrium constant, but leaves the standard potential unchanged.

The two questions below are original practice variations based on the supplied reaction. They are not additional verified PYQs.

What are the potential and equilibrium constant for the reversed reaction?

The standard potential becomes −0.46 V, and the equilibrium constant becomes the reciprocal of the original. The electron count remains two for this reversed reaction:

2 A g (s) + C u^{2 +} (a q) \to 2 A g^{+} (a q) + C u (s)

E_{r e v e r s e}^{\circ} = - 0.46 V, n = 2

K_{r e v e r s e} = \frac{1}{K} \approx \frac{1}{3.7 \times 10^{15}} \approx 2.7 \times 10^{- 16}

The negative potential and equilibrium constant below one agree. Reversal changes the direction of electron transfer, not the number transferred.

What are the electron count, potential and equilibrium constant when the equation is doubled?

The electron count becomes four, the standard potential stays at +0.46 V, and the new equilibrium constant is the square of the original. Apply those rules to:

2 C u (s) + 4 A g^{+} (a q) \to 2 C u^{2 +} (a q) + 4 A g (s)

n_{n e w} = 4, E_{n e w}^{\circ} = + 0.46 V

K_{n e w} = K^{2} \approx (3.7 \times 10^{15})^{2} \approx 1.4 \times 10^{31}

Both the electron count and logarithm double, so their ratio stays unchanged:

E_{n e w}^{\circ} = \frac{0.0591}{2 n} (2 \log_{10} K) = E^{\circ}

Both practice answers use the worked approximation, not the rounded option value. Next, apply your potential-calculation method to the NEET 2008 fuel-cell EMF worked solution.

Frequently asked questions

What is the answer to the NEET 2007 copper–silver equilibrium question?