Calculate the emf at 25 °C of the cell Zn(s) | $Z n^{2 +}$ (0.005 M) || $C u^{2 +}$ (0.5 M) | Cu(s). Standard potentials: $E_{Z n^{2 +} / Z n}^{\circ}$ = -0.76 V, $E_{C u^{2 +} / C u}^{\circ}$ = +0.34 V. Use the Nernst equation.

medium 3 marks Class 12 · Chemistry · Electrochemistry

Answer

E = 1.16 V

The correct answer is E = 1.16 V — Round to two decimal places: E $\approx$ 1.16\,V.

Step-by-step solution

Write the overall cell reaction: Zn(s) + $C u^{2 +}$ (aq) $\to$ $Z n^{2 +}$ (aq) + Cu(s).
Identify n = 2 electrons transferred.
Calculate the standard cell potential: $E_{ce l l}^{\circ} = E_{c a t h o d e}^{\circ} - E_{an o d e}^{\circ}$ =+0.34-(-0.76)=+1.10\,V.
Write the reaction quotient Q = [ $Z n^{2 +}$ ]/[ $C u^{2 +}$ ] = 0.005/0.5 = 0.01.
Apply the Nernst equation: E = E° - (0.0592/n) $lo g$ Q.
Substitute values: E = 1.10 - (0.0592/2) $lo g$ (0.01) = 1.10 - 0.0296(-2) = 1.10 + 0.0592 = 1.1592\,V.
Round to two decimal places: E $\approx$ 1.16\,V.

Common mistake: Using Q = [

C u^{2 +}

]/[

Z n^{2 +}

] instead of the correct ratio

Concept tested: Nernst equation emf calculation