Question

Calculate value of $E_{C{e}^{4+}/C{e}^{3+}}^{\circ }$ . If $E_{cell}^{\circ }$ for the cell reaction, $2C{e}^{4+}+Co\to 2C{e}^{3+}+C{o}^{2+}$ is 1.89 V. If $E_{Co/C{o}^{2+}}^{\circ }$ $=-0.28V$

• 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.

$E_{Cell}^0=E_{Cathode}^0-E_{Anode}^0=\text{1.89}=E_{Cell}^0-(-\text{0.28})$

$E_{Cell}^0=\text{1.89}-\text{0.28}=\text{1.61}$ Volts.