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REPEATER 2024 — Chemistry STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceChemistryREPEATER 20243 Marks
Question
Construct a cell using standard hydrogen electrode and zinc electrode. Write its cell reaction and cell representation.
Calculate cell potential of a cell with $0.01\ M\ Zn ^{2+}$ ions. Standard potential of a cell is $+0.76 V$
✓

Answer

Cell Reaction:
$(1)$ At the anode (zinc electrode), oxidation occurs:
$\ce{Zn (s) \longrightarrow Zn ^{2+}(aq) + 2 e^{-}}$
$(2)$ At the cathode $\text{(SHE),}$ reduction occurs:
$\ce{2 H ^{+}(aq)+2 e^{-} \longrightarrow H _2(g)}$
$(3)$ Overall Cell Reaction:
$\ce{Zn(s) + 2H ^{+}(aq) \longrightarrow Zn ^{2+}(aq) + H _2(g)}$
Cell representation:
$\ce{Zn(s)\left| Zn ^{2+}(aq) \| H ^{+}( aq )\right| H_2(g) \mid Pt(s)}$
Given: $\left[ Zn ^{2+}\right]=0.01\ M ,$
$E_{ Zn }^o=+0.76 V,$
$E_{Z n}=$ ?
Electrode reaction,
$\ce{Zn ^{2+}(aq)(0.01 M) + 2 e^{-} \longrightarrow Zn _{( s )}}$
$E_{Z n}=E_{Z n}^o-\frac{0.0592}{n} \log _{10} \frac{1}{\left[z n^{2+1}\right]}$
$\therefore \ =0.76-\frac{0.0592}{2} \log _{10} \frac{1}{0.01}$
$\therefore \ =0.76-0.0592$
$\therefore \ E_{Z n}=0.7008$

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