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Electrochemistry — Chemistry STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceChemistryElectrochemistry4 Marks
Question
Write balanced equations for the half reactions and calculate the reduction potentials at $25^{\circ} C$ for the following half cells :
$(a) \ Cl ^{-}(1.2 M ) \mid Cl _2( g , 3.6 atm ) E ^0=1.36 V$
$(b) \ Fe ^{2+}(2 M ) \mid Fe _{( s )} E ^0=-0.44 V$
✓

Answer

$(a)$ Given : Half cell,
$Cl _{( aq )}^{-}(1.2 M )\left| Cl _2( g , 3.6 atm )\right| Pt$
$ E ^0 Cl _2 / Cl ^{-}=1.36 V$
The reduction reaction:
$Cl _{2( g )}+2 e ^{-} \longrightarrow 2 Cl _{(\text {aq) }}^{-} \therefore n=2$
$ E_{ Cl _2 / Cl ^{-}}=E_{ Cl _2 / Cl ^{-}}^0-\frac{0.0592}{n} \log _{10} \frac{\left[ Cl ^{-}\right]^2}{\left[ Cl _2\right]}$
$ =1.36-\frac{0.0592}{2} \log _{10} \frac{(1.2)^2}{3.6}$
$ =1.36-0.0296(\overline{1} .6021)$
$ =1.36-0.0296(-0.3979)$
$ =1.36+0.01178$
$ =1.37178$
$ \cong 1.372 V$
$(b)$ Given: Half cell, $Fe _{( aq )}^{2+}(2 M ) \mid Fe _{( s )}$
$E ^0 Fe ^{2+} / Fe =-0.44 V$
The reduction reaction:
$Fe _{( aq )}^{2+} +2 e ^{-} \longrightarrow Fe _{( s )} \therefore n=2$
$E_{ Fe ^{2+} / Fe } =E_{ Fe ^{2+} / Fe }^0-\frac{0.0592}{n} \log _{10} \frac{[ Fe ]}{\left[ Fe ^{2+}\right]}$
$ =-0.44-\frac{0.0592}{2} \log _{10} \frac{1}{2}$
$ =-0.44-0.0296 \log _{10} 0.5$
$ =-0.44-0.0296 \times(1.6990)$
$=-0.44-0.0296 \times(-0.3010)$
$=-0.44+0.00891$
$=-0.43109 V$
$(a)$ Half reaction: $Cl _{2(g)}+2 e ^{-} \rightarrow 2 Cl _{(\text {aq) }}^{-}$$ E _{\text {Cell }}=1.372 V $
$\text { (b) } Fe _{(\text {(aq) }}^{2+}+2 e ^{-} \rightarrow Fe _{( s )}$
$E _{\text {Cell }}=-0.43109 V .$

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