If the standard emf of the cell Zn_{(s)} | Zn | vedclass

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(C) The cell reaction is: $Zn_{(s)} + 2Ag_{(1 \ M)}^{+1} \rightarrow Zn_{(1 \ M)}^{+2} + 2Ag_{(s)}$.Here, the number of electrons transferred $(n)$ is

Vedclass · Accepted Answer

$-299.150$

Vedclass · Answer

(C) The cell reaction is: $Zn_{(s)} + 2Ag_{(1 \ M)}^{+1} ightarrow Zn_{(1 \ M)}^{+2} + 2Ag_{(s)}$.Here, the number of electrons transferred $(n)$ is $2$.The standard $emf$ of the cell $(E_{cell}^o)$ is $1.55 \ V$.The electrical work done under standard conditions is equal to the change in Gibbs free energy $(\Delta G^o)$.$\Delta G^o = -nFE_{cell}^o$.Substituting the values: $n = 2$, $F = 96500 \ C \ mol^{-1}$, $E_{cell}^o = 1.55 \ V$.$\Delta G^o = -2 	imes 96500 	imes 1.55 = -299150 \ J = -299.150 \ kJ$.Therefore, the electrical work done is $-299.150 \ kJ$.

If the standard $emf$ of the cell $Zn_{(s)} | Zn_{(1 \ M)}^{+2} || Ag_{(1 \ M)}^{+1} | Ag_{(s)}$ is $1.55 \ V$, what is the electrical work done under standard conditions (in $kJ$)?

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