What is the single electrode potential of a half-cell for zinc electrode dipping in 0-01 M ZnSO4 solution of 25-C- The standard electrode potential of Zn- Zn- system is 0-763 volt at 25-C-

We know that, E_OX = E_OX^∘ 0.0591nlog_10[ion] Putting the value of E_OX^∘ = 0.763 V, n = 2 and [Zn^2+] = 0.01 M , E_OX = 0.763 0.05912l ...

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What is the single electrode potential of a half-cell for zinc electrode dipping in $0.01 M Z n S O_{4}$ solution of $25^{\circ} C$ ? The standard electrode potential of $Z n / Z n^{+}$ system is $0.763$ volt at $25^{\circ} C$ .

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Solution

We know that, $E_{O X} = E_{O X}^{\circ} - \frac{0.0591}{n} {log}_{10} [i o n]$
Putting the value of $E_{O X}^{\circ} = 0.763 V, n = 2$ and $[Z n^{2 +}] = 0.01 M$ ,
$E_{O X} = 0.763 - \frac{0.0591}{2} {log}_{10} [0.01]$
$= 0.763 - 0.02955 \times (- 2)$
$= (0.763 + 0.0591) v o l t = 0.8221 v o l t$ .

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What is the single electrode potential of a half-cell for zinc electrode dipping in 0.01 M ZnSO4 solution of 25∘C? The standard electrode potential of Zn/Zn+ system is 0.763 volt at 25∘C.

We know that, EOX=E∘OX−0.0591nlog10[ion]Putting the value of E∘OX=0.763 V,n=2 and [Zn2+]=0.01 M,EOX=0.763−0.05912log10[0.01]=0.763−0.02955×(−2)=(0.763+0.0591)volt=0.8221 volt.

What is the single electrode potential of a half-cell for zinc electrode dipping in $0.01 M Z n S O_{4}$ solution of $25^{\circ} C$ ? The standard electrode potential of $Z n / Z n^{+}$ system is $0.763$ volt at $25^{\circ} C$ .