For the galvanic cell,
Zn(s) + Cu²⁺ (0.02 M) → Zn²⁺ (0.04 M) + Cu(s),
E cell = ______________ × 10⁻² V. (Nearest integer)
[Use : E_(Cu/Cu²⁺)⁰ = - 0.34 V, E_(Zn/Zn²⁺)⁰ = + 0.76 V, 2.303RT/F = 0.059 V]

Solution

Chemistry ma’am · 4 steps

The data gives oxidation potentials, so flip their signs first, then let Nernst trim the 1.10 V.

1. Convert to reduction. The given values are oxidation potentials, so reverse the signs. E⁰_(Cu²⁺/Cu) = +0.34 V and E⁰_(Zn²⁺/Zn) = -0.76 V.
2. Standard cell potential. Zn is oxidised at the anode and Cu²⁺ is reduced at the cathode. So E⁰_(cell) = 0.34 - (-0.76) = 1.10 V, with n = 2.

   E_(cell) = E⁰_(cell) - 0.059/n log[Zn²⁺]/[Cu²⁺]

3. Put in numbers. The ratio is 0.04/0.02 = 2, and log 2 = 0.301. So E_(cell) = 1.10 - 0.059/2 × 0.301 = 1.10 - 0.0089 = 1.0911 V.
4. So. E_(cell) = 109.1 × 10⁻² V, which rounds to 109.

The common slip: 111. Putting [Cu²⁺] on top of the ratio adds 0.0089 V instead of subtracting it. That gives 110.9 × 10⁻² V, which rounds to 111, not 109.

The idea to keep. Always check whether the given E⁰ values are oxidation or reduction potentials before using them. In the log term, the products go on top and the reactants go below, with solids left out.

Answer from the NTA final answer key. Solution by PrepWiser.