tch-001/c85898bcfb8f.webp" data-orsrc="https://uploads-pracbee-ramanuj.s3.ap-south-1.amazonaws.com/pyq-question-images/batch-001/c85898bcfb8f.webp" loading="lazy" style="max-width:100%;height:auto;display:block;margin:0 auto;max-height:40vh;vertical-align:baseline" alt="AP EAPCET 2021 - 19th August Evening Shift Chemistry - Electrochemistry Question 32 English Explanation 1">

Anode $$\mathrm{Cr} \longrightarrow \mathrm{Cr}^{3+}+3 e^{-}$$ ..... (i)

Cathode $$\mathrm{Fe}^{2+} \longrightarrow 2 e^{-}+\mathrm{Fe}$$ ..... (ii)

Multiply 2 in Eq. (i) and 3 in Eq. (ii) we get,

AP EAPCET 2021 - 19th August Evening Shift Chemistry - Electrochemistry Question 32 English Explanation 2

For, $$E_{\text {cell }}^{\circ}=E_{\text {cathode }}^{\circ}-E_{\text {anode }}^{\circ}$$

$$=-0.42-(-0.72 \mathrm{~V})=-0.42+0.72 \mathrm{~V}$$

$$E_{\text {cell }}^{\circ}=0.30 \mathrm{~V}$$

Using Nernst equation,

$$E_{\text {cell }}=E_{\text {cell }}^{\circ}-\frac{0.0591}{n} \log \frac{\left[\mathrm{Cr}^{3+}\right]^2}{\left[\mathrm{Fe}^{2+}\right]^3}$$

Here, $$n=6\left(\right.$$ For $$\left.6 e^{-}\right)$$

$$\begin{aligned} & E_{\text {cell }}=0.30-\frac{0.0591}{6} \log \frac{(0.1)^2}{(0.1)^3} \\ & E_{\text {cell }}=0.30-0 \Rightarrow E_{\text {cell }}=0.30 \mathrm{~V} \end{aligned}$$

Hence, emf of the reaction cell is $$0.30 \mathrm{~V}$$.

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