The cell voltage of Daniel cell [Zn |ZnSO₄(aq) | CuSO₄(aq) | Cu] is 1.07 V. If reduced potential of Cu²⁺ Cu is 0.34 V, the reduction potential of Zn²⁺ Zn is

(d) -0.73 V. Explanation: Ecell = E(Cu) − E(Zn) ⇒ E(Zn²⁺/Zn) = 0.34 − 1.07 = −0.73 V.

Q94 · CSIR-NET Chemistry, December 2013

Paper: CSIR-NET December 2013 · Subject: Physical Chemistry · Chapter: Electrochemistry · Topic: Electrochemistry – General · Marks: 2 · Difficulty: Medium

The cell voltage of Daniel cell $\left[\mathrm{Zn}\left|\mathrm{ZnSO}_{4}(\mathrm{aq}) \| \mathrm{CuSO}_{4}(\mathrm{aq})\right| \mathrm{Cu}\right]$ is 1.07 V. If reduced potential of $\mathrm{Cu}^{2+} \mid \mathrm{Cu}$ is 0.34 V, the reduction potential of $\mathrm{Zn}^{2+} \mid \mathrm{Zn}$ is

(a)1.141 V

(b)-1.41 V

(c)0.73 V

(d)-0.73 V

Answer

Answer: D ✓ checked by 4AB · confidence high

Explanation

E_cell = E(Cu) − E(Zn) ⇒ E(Zn²⁺/Zn) = 0.34 − 1.07 = −0.73 V.

Study loop for Electrochemistry

1. Practise the PYQsPrevious-year questions, with answers

· Browse this chapter in the app