A zinc electrode is placed in a $0.1$ M solution at $25^{o}$ C. Assuming that the salt is $20 %$ dissociated at this dilution, calculate the electrode potential. $E^{0} (Z n^{2 +} | Z n) = - 0.76$ V. $[l o g 2 = 0.3]$

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Solution

$20$ % of $0.1$ M solution dissociated.
$∴$ Concentration of ${Z n}^{+ 2}$ ions in ${Z n S O}_{4}$ solution $= \frac{0.1 \times 20}{100} = 0.02$ M
now the electrode will be : $Z n (s) / {Z n}^{+ 2} (0.02 M)$
${Z n}^{+ 2} + 2 e^{-} \to Z n (s)$
$E_{{Z n}^{2 +} / Z n} = E_{{Z n}^{+ 2} / Z n}^{0} - \frac{0.059}{2} l o g \frac{1}{0.02} [x = 2]$
$= - 0.76 + \frac{0.059}{2} l o g 0.02 = 0.81$ volt.

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A zinc electrode is placed in a 0.1M solution at 25oC. Assuming that the salt is 20% dissociated at this dilution, calculate the electrode potential. E0(Zn2+|Zn)=−0.76V. [log2=0.3]

20% of 0.1M solution dissociated.∴ Concentration of Zn+2 ions in ZnSO4 solution =0.1×20100=0.02Mnow the electrode will be : Zn(s)/Zn+2(0.02M)Zn+2+2e−→Zn(s)EZn2+/Zn=E0Zn+2/Zn−0.0592log10.02[x=2] =−0.76+0.0592log0.02=0.81volt.

A zinc electrode is placed in a $0.1$ M solution at $25^{o}$ C. Assuming that the salt is $20 %$ dissociated at this dilution, calculate the electrode potential. $E^{0} (Z n^{2 +} | Z n) = - 0.76$ V. $[l o g 2 = 0.3]$