Annette Sabin

## Answered question

2022-01-17

Calculate the standard potential, ${E}^{\circ }$, for this reaction from its $\mathrm{△}{G}^{\circ }$ value.
$X\left(s\right)+Y3+\left(aq\right)⇒X3+\left(aq\right)+Y\left(s\right)$

### Answer & Explanation

jgardner33v4

Beginner2022-01-18Added 35 answers

Step 1
are related as $÷$

where n is no. of electrons transferred.
Step 2
Given, $\mathrm{△}{G}^{\circ }=-28.0kj$
${E}^{\circ }=?$
$x+{y}^{3+}⇒{x}^{3+}+y$
Reaction at anade :
$x⇒{x}^{3+}+3{e}^{-}$
$\therefore n=3$

$=\frac{-\left(-28.0\right)kJ}{3×96500C}$
$=9.67×{10}^{-5}\frac{kJ}{c}$
$=9.67×{10}^{-5}×{10}^{3}\frac{J}{c}=0.0967V$
Answer: 0.0967V

Samantha Brown

Beginner2022-01-19Added 35 answers

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

alenahelenash

Skilled2022-01-23Added 366 answers

Answer: 0.124 V Explanation: $X\left(s\right)+{Y}^{4+}\left(aq\right)⇒{X}^{4+}\left(aq\right)+Y\left(s\right)$ To calculate standard Gibbs free energy, we use the equation: 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: $-48.0×1000=-4×96500×{E}_{cell}^{o}$ ${E}_{cell}^{o}=0.124$ Thus the standard potential ${E}^{\circ }$, for this reaction is 0.124 V

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