From the standard potential in acidic medium as shown in the following la-timer diagram, the value of $({E_{1}}^{0} + {E_{2}}^{0})$ . in volts is:

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Solution

Comparing gibb's free energy,
$n F E^{0} = n_{1} F E_{0}^{1} + n_{2} F E_{0}^{2}$
$3 \times E^{0} = 0.55 + 2 \times 2.26$
$E^{0} = \frac{5.07}{3} = 1.69 V \to (I)$
comparing gibb's free energy,
$n F E^{0} = n F E_{1} + n F E_{2}$
$2 \times E^{0} = 1 \times 0.96 + 1 \times 1.66$
$E^{0} = 1.31 V \to (I I)$
Adding $(I)$ and $(2)$ , $1.69 + 1.31 = 3.0 V$

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