Welcome to the Quiz!

This quiz contains 11 questions from a mix of 1 subtopics.

A reaction will take place if the redox system of the species being oxidised has a more E value than the redox system of the species being reduced.

positive

0

/

1

Do feasible redox reactions have Ecell values that are positive or negative?

negative

positive

0

/

1

Which reaction between chromium and lead is feasible?

Cr3+(aq) + e- ⇌ Cr2+(aq)    E = -0.41 V

Pb2+(aq) + 2e- ⇌ Pb(s)      E = -0.76 V

2Cr3+ + Pb ➔ 2Cr2+ + Pb2+

2Cr2+ + Pb2+ ➔ 2Cr3+ + Pb

0

/

1

Which redox reaction is the most feasible?

Cr3+(aq) + 3e- ⇌ Cr(s)      E = -0.77 V

Cu2+(aq) + 2e- ⇌ Cu(s)     E = +0.34 V

Ag+(aq) + e- ⇌ Ag(s)        E = +0.80 V

2Ag+ + Cu ➔ 2Ag + Cu2+

3Cu2+ + 2Cr ➔ 3Cu + 2Cr3+

3Ag+ + Cr ➔ 3Ag + Cr3+

0

/

1

Calculate the value of Ecell for the reaction of Mg with Zn2+ ions:

Mg2+(aq) + 2e- ⇌ Mg(s)    E = -2.38 V

Zn2+(aq) + 2e- ⇌ Zn(s)      E = -0.76 V

1.62

V

V

0

/

1

What is the correct formula for ΔG?

ΔG = -nFEcell

ΔG = nEcell

ΔG = nFEcell

ΔG = -nEcell

0

/

1

Calculate the value of ΔG for the following reaction:

Mg(s) + Zn2+(aq) ➔ Mg2+(aq) + Zn(s)    Ecell = +1.62 V

The Faraday, F = 96,500 C mol-1

Give your answer to 3 significant figures.

-313

kJ mol-1

kJ mol-1

0

/

1

How would Cu/Cu2+ elecrode potential (E) change if the concentration of Cu2+ ions was lower than 1 mol dm-3?

Cu2+(aq) + 2e- ⇌ Cu(s)     E = +0.34 V

E would become more positive

E would remain unchanged

E would become less positive

0

/

1

How would Fe2+/Fe3+ elecrode potential, E, change if the concentrations of Fe2+ and Fe3+ ions were changed by the same amount?

Fe3+(aq) + e- ⇌ Fe2+(aq)     E = +0.77 V

E would become more positive

E would become less positive

E would remain unchanged

0

/

1

What is the correct formula of the Nerst equation at 298 K?

E=E+z0.059log10[oxidised species][reduced species]
E=E+0.059zlog10[oxidised species][reduced species]
E=E+0.059zlog10[reduced species][oxidised species]
E=E+z0.059log10[reduced species][oxidised species]

0

/

1

Calculate the value of the electrode potential, E, at 298 K of a V3+(aq)/V2+(aq) electrode that has a concentration of V3+ ions of 0.180 mol dm-3 and a concentration of V2+ ions of 0.900 mol dm-3.

V3+(aq) + e- ⇌ V2+(aq)     E = -0.26 V

+0.44 V

-0.30 V

-0.22 V

-0.96 V

0

/

1