Question

Calculate the electrode potentials of the following half-cells.
(a) ${\mathrm{Ag}}^{+}(0.0436\mathrm{M})\mid \mathrm{Ag}$
(b) ${\mathrm{Fe}}^{3+}\left(5.34\times {10}^{-4}\mathrm{M}\right),{\mathrm{Fe}}^{2+}(0.090\mathrm{M})\mid \mathrm{Pt}$
(c) $\mathrm{AgBr}(sat\text{'}d),{\mathrm{Br}}^{-}(0.037\mathrm{M})\mid \mathrm{Ag}$

Short answer

(a)The electrode potential for the half cell is $0.705\mathrm{V}$.

(b) The electrode potential for the half cell is $0.642\mathrm{V}$.

(c) The electrode potential for the half cell is$0.150\mathrm{V}$.

Step-by-step solution

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(a) The half cell reaction is as follows:

${\mathrm{Ag}}^{+}+{\mathrm{e}}^{-}\rightleftharpoons \mathrm{Ag}{E}^0=0.799\mathrm{V}$
The formula for determining $E_{\mathrm{Ag}}$for the half reaction is as follows:
$E_{\mathrm{Ag}}=E_{\mathrm{Ag}}°-\frac{0.0592}{n}\log \frac{1}{\left[{\mathrm{Ag}}^{+}\right]}$
In the above equation, $n$ is the number of moles of electrons involved in the reaction.
Substitute the given value for concentration of ${\mathrm{Ag}}^{+}(0.0261\mathrm{M})$and the value of $E_{\mathrm{Ag}}°$to determine the value of $E_{\mathrm{Ag}}$.
$E_{\mathrm{Ag}}=E_{\mathrm{Ag}}^o-\frac{0.0592}{1}\log \frac{1}{\left[{\mathrm{Ag}}^{+}\right]}$
$=+0.799-\frac{0.0592}{1}\log \frac{1}{0.0261}$
$=0.705\mathrm{V}$
Hence, the electrode potential for the half cell is $0.705\mathrm{V}.$

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(b) The half cell reaction is as follows:
${\mathrm{Fe}}^{3+}+{\mathrm{e}}^{-}\rightleftharpoons {\mathrm{Fe}}^{2+}{E}^0=0.771\mathrm{V}$
The formula for determining $E_{\text{Ag}}$ for the half reaction is as follows:

$E_{{\mathrm{Fe}}^{2*}}=E_{{\mathrm{Fe}}^{2*}}°-\frac{0.0592}{n}\log \frac{\left[{\mathrm{Fe}}^{2+}\right]}{\left[{\mathrm{Fe}}^{3+}\right]}$
In the above equation, $n$ is the number of moles of electrons involved in the reaction.
Substitute the given values for concentration and the value of $E_{{\mathrm{Fe}}^{2+}}°$to determine the value of $E_{{\mathrm{Fe}}^{2+}}$

$E_{{\mathrm{Fe}}^{2*}}=E_{{\mathrm{Fe}}^{2*}}°-\frac{0.0592}{n}\log \frac{\left[{\mathrm{Fe}}^{2+}\right]}{\left[{\mathrm{Fe}}^{3+}\right]}$$=+0.771-\frac{0.0592}{1}\log$$\frac{0.100}{6.72\times {10}^{-4}}$
$=0.642\mathrm{V}$
Hence, the electrode potential for the half cell is $0.642\mathrm{V}$

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(c) The half cell reaction is as follows:
$\mathrm{AgBr}\left(s\right)+{\mathrm{e}}^{-}\rightleftharpoons \mathrm{Ag}\left(s\right)+{\mathrm{Br}}^{-}{E}^0=0.073\mathrm{V}$
The formula for determining $E_{\text{Ag}}$ for the half reaction is as follows:
$E_{{\mathrm{Br}}^{-}}=E_{{\mathrm{Br}}^{-}}^o-\frac{0.0592}{n}\log \left[{\mathrm{Br}}^{-}\right]$
In the above equation, $n$ is the number of moles of electrons involved in the reaction.
Substitute the given value for concentration of $(0.050\mathrm{M})$and the value of $E_{\mathrm{Ag}}°$to determine the value of $E_{\mathrm{Ag}}$
$E_{{\mathrm{Br}}^{-}}$$=E_{{\mathrm{Br}}^{-}}°-\frac{0.0592}{1}\log \left[{\mathrm{Br}}^{-}\right]$
$=+0.073-\frac{0.0592}{1}\log 0.050$
$=0.150\mathrm{V}$
Hence, the electrode potential for the half cell is $0.150\mathrm{V}.$