A student wants to measure the standard electrode potential of the Ni^2+(aq)/Ni(s) half-cell. They set up a nickel electrode dipping into 1.00 mol dm^-3 Ni^2+(aq) solution, connected by a salt bridge to a standard hydrogen electrode, with a high-resistance voltmeter completing t…

Answer: ((a)) 298 K; H2(g) at 100 kPa (1 atm); [H^+(aq)] = 1.00 mol dm^-3; a platinised (platinum black coated) platinum electrode ((b)) Nickel electrode is negative. Nickel electrode (oxidation): Ni(s) → Ni^2+(aq) + 2e^-. Hydrogen electrode (reduction): 2H^+(aq) + 2e^- → H2(g) ((c)) E^ (Ni^2+/Ni) = E^ (cathode) - Ecell^ = 0.00 - 0.25 = -0.25 V ((d)) Ni(s) Ni^2+(aq) ∥ H^+(aq) H2(g) Pt(s). Part (a): Standard hydrogen electrode conditions For the platinum/hydrogen electrode to be a valid standard hydrogen electrode, it must be set up under standard conditions: - temperature of 298 K - H2(g) bubbled over the electrode at a pre…

A2 Chemistry 9701: Measuring E for Ni2+/Ni Against the Standard Hydrogen Electrode

A student wants to measure the standard electrode potential of the $\text{Ni}^{2+}(aq)/\text{Ni}(s)$ half-cell. They set up a nickel electrode dipping into $1.00\ \text{mol dm}^{-3}\ \text{Ni}^{2+}(aq)$ solution, connected by a salt bridge to a standard hydrogen electrode, with a high-resistance voltmeter completing the external circuit. At $298\ \text{K}$ , the voltmeter reads $0.25\ \text{V}$ , and electrons are found to flow through the external wire from the nickel electrode to the platinum/hydrogen electrode.

(a) State the conditions required for the platinum/hydrogen electrode to function as a standard hydrogen electrode. [2]

(b) State which electrode is negative, and write the half-equation occurring at each electrode. [3]

(c) Given that $E^{\ominus}$ of the standard hydrogen electrode is defined as $0.00\ \text{V}$ , calculate the standard electrode potential, $E^{\ominus}$ , of the $\text{Ni}^{2+}(aq)/\text{Ni}(s)$ half-cell. [2]

(d) Write the cell notation for this electrochemical cell. [2]

Part (a): Standard hydrogen electrode conditions

For the platinum/hydrogen electrode to be a valid standard hydrogen electrode, it must be set up under standard conditions:

temperature of $298\ \text{K}$
$\text{H}_2(g)$ bubbled over the electrode at a pressure of $100\ \text{kPa}$ (1 atm)
$[\text{H}^+(aq)] = 1.00\ \text{mol dm}^{-3}$
an inert platinum electrode coated in finely divided platinum (“platinum black”), which adsorbs $\text{H}_2(g)$ and provides a large surface area so the equilibrium $2\text{H}^+(aq) + 2e^- \rightleftharpoons \text{H}_2(g)$ is established rapidly

Part (b): Polarity and half-equations

Electrons flow through the external circuit from the nickel electrode to the hydrogen electrode. An electrode that releases electrons into the external circuit is the site of oxidation, and by convention this is the negative electrode.

Nickel electrode. Negative, oxidation: $\text{Ni}(s) \rightarrow \text{Ni}^{2+}(aq) + 2e^-$
Hydrogen electrode. Positive, reduction (it receives the electrons released by nickel): $2\text{H}^+(aq) + 2e^- \rightarrow \text{H}_2(g)$

Part (c): Calculating $E^{\ominus}(\text{Ni}^{2+}/\text{Ni})$

Since the hydrogen electrode is the cathode (reduction) and the nickel electrode is the anode (oxidation):

$E_{cell}^{\ominus} = E^{\ominus}(\text{cathode}) - E^{\ominus}(\text{anode}) = E^{\ominus}(\text{H}^+/\text{H}_2) - E^{\ominus}(\text{Ni}^{2+}/\text{Ni})$

Substituting $E_{cell}^{\ominus} = 0.25\ \text{V}$ (the voltmeter reading) and $E^{\ominus}(\text{H}^+/\text{H}_2) = 0.00\ \text{V}$ by definition:

$0.25 = 0.00 - E^{\ominus}(\text{Ni}^{2+}/\text{Ni})$

$E^{\ominus}(\text{Ni}^{2+}/\text{Ni}) = 0.00 - 0.25 = -0.25\ \text{V}$

Check: a negative $E^{\ominus}$ for nickel is consistent with the observation that nickel is the electrode that is oxidised (it releases electrons more readily than the $\text{H}^+/\text{H}_2$ couple, which corresponds to a smaller, more negative electrode potential), the sign and the observed direction of electron flow agree.

Part (d): Cell notation

By convention, the anode (oxidation, negative electrode) is written on the left and the cathode (reduction, positive electrode) on the right, separated by the salt bridge (||):

$\text{Ni}(s) \mid \text{Ni}^{2+}(aq) \parallel \text{H}^+(aq) \mid \text{H}_2(g) \mid \text{Pt}(s)$

Final answers

(a) $298\ \text{K}$ ; $\text{H}_2(g)$ at $100\ \text{kPa}$ ; $[\text{H}^+(aq)] = 1.00\ \text{mol dm}^{-3}$ ; platinised platinum electrode.
(b) Nickel electrode is negative (oxidised): $\text{Ni}(s) \rightarrow \text{Ni}^{2+}(aq) + 2e^-$ . Hydrogen electrode (reduced): $2\text{H}^+(aq) + 2e^- \rightarrow \text{H}_2(g)$ .
(c) $E^{\ominus}(\text{Ni}^{2+}/\text{Ni}) = \boxed{-0.25\ \text{V}}$
(d) $\text{Ni}(s) \mid \text{Ni}^{2+}(aq) \parallel \text{H}^+(aq) \mid \text{H}_2(g) \mid \text{Pt}(s)$