Deduce a balanced equation for the overall reaction when the standard nickel and iodine half-cells are connected.

Predict, giving a reason, the direction of movement of electrons when the standard nickel and manganese half-cells are connected.

Calculate the cell potential, in V, when the standard iodine and manganese half-cells are connected.

Identify the best reducing agent in the table above.

State and explain the products of electrolysis of a concentrated aqueous solution of sodium chloride using inert electrodes. Your answer should include half-equations for the reaction at each electrode.

Ni (s) + I₂ (aq) → 2I^– (aq) + Ni²⁺ (aq)

electron movement «in the wire» from Mn(s) to Ni(s)

E^θ «for reduction» of Ni²⁺ is greater/less negative than E^θ «for reduction» of Mn²⁺

OR

Ni²⁺ is stronger oxidizing agent than Mn²⁺

OR

Mn is stronger reducing agent than Ni

«0.54 V – (–1.18 V) = +»1.72 «V»

Do not accept –1.72 V.

Mn «(s)»

Positive electrode (anode):
2Cl^– (aq) → Cl₂ (g) + 2e^–

Cl^– oxidized because higher concentration

OR

electrode potential/E depends on concentration

OR

electrode potential values «of H₂O and Cl^–» are close

Negative electrode (cathode):
2H₂O (l) + 2e^– → H₂ (g) + 2OH^– (aq)

OR

2H⁺ (aq) + 2e^– → H₂ (g)

H₂O/H⁺ reduced because Na⁺ is a weaker oxidizing agent

OR

Na⁺ not reduced to Na in water

OR

H⁺ easier to reduce than Na⁺
OR

H lower in activity series «than Na»

Accept \( \rightleftharpoons \).