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