Question

Calculate value of $E^\circ _{C e^{4 +} / C e^{3 +}}$ . If $E^\circ _{c e l l}$ for the cell reaction, $2Ce^{4 +}+Co \rightarrow 2Ce^{3 +}+Co^{2 +}$ is 1.89 V. If $E^\circ _{C o / C o^{2 +}}$ $= \, -0.28 \, V$

• A. $- \, 1.64V$
• B. $+ \, 1.64V$
• C. $- \, 2.08V$
• D. $+ \, 2.17V$

Answer 1

Answer: (B)

In this cell $Co$ is oxidised and it acts as anode and Ce acts as cathode.

$E_{C e l l}^{0}=E_{C a t h o d e}^{0}-E_{A n o d e}^{0}=\text{1.89}=E_{C e l l}^{0}-\left(\right.-\text{0.28}\left.\right)$

$E_{C e l l}^{0}=\text{1.89}-\text{0.28}=\text{1.61}$ Volts.