Emf of the following cell at 298K in V is x × 10⁻².
Zn|Zn²⁺(0.1 M)||Ag⁺ (0.01 M)|Ag
The value of x is _________. (Rounded off to the nearest integer)
[Given : E_(Zn²⁺/Zn)^θ = - 0.76V;E_(Ag²⁺/Ag)^θ = + 0.80V;2.303RT/F = 0.059]

Solution

Chemistry ma’am · 4 steps

Balance the cell reaction first: two silver ions for each zinc, so n = 2 and [Ag⁺] gets squared.

1. Cell reaction. Zinc on the left is oxidised and silver on the right is reduced: Zn + 2Ag⁺ → Zn²⁺ + 2Ag. Two electrons move, so n = 2. The data line says Ag²⁺, but the cell uses Ag⁺, so read it as the Ag⁺/Ag potential.
2. Standard emf. E°_(cell) = E°_(cathode) - E°_(anode) = 0.80 - (-0.76) = 1.56 V.

   E_(cell) = E°_(cell) - 0.059/n log[Zn²⁺]/[Ag⁺]²

3. Put in numbers. [Zn²⁺]/[Ag⁺]² = 0.1/(0.01)² = 0.1/10⁻⁴ = 10³, so log 10³ = 3. Then E_(cell) = 1.56 - 0.059/2 × 3 = 1.56 - 0.0885 = 1.4715 V.
4. So. E_(cell) = 147.15 × 10⁻² V, so x rounds to 147.

The common slip: 153. Leaving [Ag⁺] unsquared makes the ratio 10 instead of 1000, so E_(cell) = 1.5305 V and x comes out as 153, not 147.

The idea to keep. In the Nernst equation, every concentration is raised to its coefficient in the balanced cell reaction, and n is the number of electrons in that same balanced equation.

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