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: \(E_{cell} \approx 0.819\) V

Construction: A zinc rod dipped in a Zn²⁺ solution (ZnSO₄) is the anode. The standard hydrogen electrode (platinised Pt foil dipped in 1 M H⁺, with pure H₂ gas bubbled at 1 bar) is the cathode. The two solutions are joined by a salt bridge and the electrodes by an external wire through a voltmeter. Electrons flow from zinc to the SHE.

Reactions: Anode: Zn_(s) → Zn²⁺_(aq) + 2e⁻; Cathode: 2H⁺_(aq) + 2e⁻ → H_2(g)

Cell reaction: Zn_(s) + 2H⁺_(aq) → Zn²⁺_(aq) + H_2(g)

Cell representation: Zn_(s) | Zn²⁺ (0.01 M) || H⁺ (1 M) | H₂ (g, 1 bar) | Pt

Cell potential (\(n = 2\), \([H^+] = 1\) M, \(P_{H_2} = 1\) bar):

\(E_{cell} = E^0_{cell} - \dfrac{0.0592}{n}\log\dfrac{[Zn^{2+}]P_{H_2}}{[H^+]^2} = 0.76 - \dfrac{0.0592}{2}\log(0.01)\)

\(= 0.76 - 0.0296\times(-2) = 0.76 + 0.0592 = 0.8192\) V