Calculate the standard potential, E∘, for this reaction from its △G∘ value. X(s)+Y3+(aq)⇒X3+(aq)+Y(s) △?∘=−28.0 kJ

Step 1
${\displaystyle E^{\circ }\text{and}\mathrm{△}{G}^{\circ }}$ are related as ${\displaystyle ÷}$
${\displaystyle \mathrm{△}{G}^{\circ }=-nF{E}^{\circ }}$
where n is no. of electrons transferred.
Step 2
Given, ${\displaystyle \mathrm{△}{G}^{\circ }=-28.0kj}$
${\displaystyle E^{\circ }=?}$
${\displaystyle x+y^{3+}\Rightarrow x^{3+}+y}$
Reaction at anade :
${\displaystyle x\Rightarrow x^{3+}+3e^{-}}$
${\displaystyle \therefore n=3}$
${\displaystyle \mathrm{△}{G}^{\circ }=-nF{E}^{\circ }}$
${\displaystyle E^{\circ }=\frac{-\mathrm{△}{G}^{\circ }}{nF}}$
${\displaystyle =\frac{-(-28.0)kJ}{3\times 96500C}}$
${\displaystyle =9.67\times {10}^{-5}\frac{kJ}{c}}$
${\displaystyle =9.67\times {10}^{-5}\times {10}^3\frac{J}{c}=0.0967V}$
Answer: 0.0967V

Step 1
The reaction given is
${\displaystyle X\left(s\right)+Y4+\left(aq\right)\Rightarrow X4+\left(aq\right)+Y\left(s\right)}$
In the above balanced reaction, 1 molecule of X (s) is being oxidised from ${\displaystyle 0\to 4+}$ oxidation state.
Hence there are total 4 electrons required for the oxidation of X (s) from ${\displaystyle 0\to 4+}$.
Hence in the reaction, total 4 electrons are transferred.
Step 2
The relationship between the $E_{cell}^{0}and\mathrm{△}{G}^0$ is given by
$nF{E}_{cell}^0=-\mathrm{△}{G}^0$
where ${\displaystyle n=\text{number of electrons transferred in the reaction}=4}$
${\displaystyle F=\text{Faraday's constant}=96500C}$
and ${\displaystyle \mathrm{△}{G}^0=91.0KJ=91000J}$ (since ${\displaystyle 1KJ=1000J}$)
Hence substituting the values we get
${\displaystyle 4\times 96500\times E_{cell}^0=-91000}$
${\displaystyle \Rightarrow E_{cell}^0=-0.23575V\approx }$

Answer: 0.124 V Explanation: $X(s)+Y^{4+}(aq)\Rightarrow X^{4+}(aq)+Y(s)$ To calculate standard Gibbs free energy, we use the equation: $\mathrm{△}{G}^{\circ }=-nF{E}_{cell}^o$ Where, $n=\text{number of electrons transferred}=4$ $F=\text{Faradays constant}=96500C$ $E_{cell}^o=\text{standard cell potential}=?$ Putting values in above equation, we get: $-48.0\times 1000=-4\times 96500\times E_{cell}^o$ $E_{cell}^o=0.124$ Thus the standard potential $E^{\circ }$, for this reaction is 0.124 V