chapter 5 electrochemistry class 12 chemistry textbook solution

3. Answer the following in brief

v. Construct a galvanic cell from the electrodes
| Co and Mn2+/Mn. E_Co^circ` = 1.82 V, `”E”_”Mn”^circ` = –1.18 V. Calculate `”E”_”cell”^circ`

Answer:-

Given: \[E^\circ_{\text{Co}} = 1.82 \, \text{V},\] \[E^\circ_{\text{Mn}} = -1.18 \, \text{V}.\] To find: \(E^\circ_{\text{cell}}\) and cell representation Formulae: \[E^\circ_{\text{cell}} = E^\circ_{\text{Cathode}} - E^\circ_{\text{Anode}}\] Calculation: Electrode reactions are At anode: \(3\text{Mn}_{(s)} \rightarrow \text{Mn}_{(aq)}^{2+} + 2e^-\) At cathode: \(2\text{Co}_{(aq)}^{3+} + 3e^- \rightarrow 2\text{Co}_{(s)}\) The cell is composed of Mn (anode), Mn\(_{(s)} | \text{Mn}_{(aq)}^{2+}\) and Co (cathode), \(\text{Co}_{(aq)}^{3+} | \text{Co}_{(s)}\) The cell is represented as: \(\text{Mn}_{(s)} | \text{Mn}_{(aq)}^{2+} | \text{Co}_{(aq)}^{3+} | \text{Co}_{(s)}\) The standard electrode potential is given by: \[E^\circ_{\text{cell}} = E^\circ_{\text{Cathode}} - E^\circ_{\text{Anode}}\] \[= 1.82 \, \text{V} - (-1.18 \, \text{V})\] \[= 3.00 \, \text{V}\] The standard cell potential is \(3.00 \, \text{V}\).

chapter 5 elctrochemistry textbook solution page 118