Question 17N.2.hl.TZ0.7
| Date | November 2017 | Marks available | [Maximum mark: 9] | Reference code | 17N.2.hl.TZ0.7 |
| Level | hl | Paper | 2 | Time zone | TZ0 |
| Command term | Calculate, Deduce, Identify, Predict, State and explain | Question number | 7 | Adapted from | N/A |
Consider the following half-cell reactions and their standard electrode potentials.
Deduce a balanced equation for the overall reaction when the standard nickel and iodine half-cells are connected.
[1]
Ni (s) + I2 (aq) → 2I– (aq) + Ni2+ (aq)
Predict, giving a reason, the direction of movement of electrons when the standard nickel and manganese half-cells are connected.
[2]
electron movement «in the wire» from Mn(s) to Ni(s)
Eθ «for reduction» of Ni2+ is greater/less negative than Eθ «for reduction» of Mn2+
OR
Ni2+ is stronger oxidizing agent than Mn2+
OR
Mn is stronger reducing agent than Ni
Calculate the cell potential, in V, when the standard iodine and manganese half-cells are connected.
[1]
«0.54 V – (–1.18 V) = +»1.72 «V»
Do not accept –1.72 V.
Identify the best reducing agent in the table above.
[1]
Mn «(s)»
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.
[4]
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 .