Q13 / 50 · Manipal (MET) 2023 · Chemistry
Value of $E_2^0$ is $\qquad$ in the given diagram (ignore -ve sign)
Correct answer: 0.77
Solution
$\Delta G_3=\Delta G_1+\Delta G_1$
$n_3 F E_3^{\circ}=n_1 F E_1^{\circ}+n_2 F E_2^{\circ}$
$3 \times 0.036=2 \times 0.439+1 \times E_2^{\circ}$
$\Rightarrow \quad E_2^{\circ}=0.108-0.878$
$=-0.77 \mathrm{~V}$
Hence, the correct answer is 0.77.