Determine whether the following redox reactions are spontaneous as written? (Use the information in the table below. Show the half reactions and the potentials for each half reaction.)
a. 2Ag(s) + Mg2+(aq) → 2Ag+ (aq) + Mg(s) [5 points]
b. 2Al(s) + 3Zn2+(aq) → 2Al3+(aq) + 3Zn(s) [5 points]
Reduction Potentials at 25°C
Electrode Half-Reaction E0(V)
Mg2+/Mg Mg2+ + 2e– → Mg –2.37
Al3+/Al Al3+ + 3e– → Al –1.66
Zn2+/Zn Zn2+ + 2e– → Zn –0.76
Ag2+/Ag Ag+ + e– → Ag +0.80
2Ag(s) + Mg2+(aq) → 2Ag+ (aq) + Mg(s) [5 points]
2Ag(s) ==>2 Ag^+ + e ...........E = -0.80
Mg^2+ + 2e ==> Mg(s) ......E = -2.87
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2Ag(s) + Mg^2+ ==> 2Ag^+ +Mg(s) E = -3.67
E cell is negative. It will not be spontaneous.
The others are done the same way.
To determine whether a redox reaction is spontaneous as written, we need to compare the reduction potentials of the half-reactions involved.
Let's break down the given reactions into their respective half-reactions and find the potentials for each:
a. 2Ag(s) + Mg2+(aq) → 2Ag+(aq) + Mg(s)
The half-reactions involved are:
Reduction: Mg2+ + 2e– → Mg (E0 = –2.37 V)
Oxidation: 2Ag → 2Ag+ + 2e– (E0 = +0.80 V)
b. 2Al(s) + 3Zn2+(aq) → 2Al3+(aq) + 3Zn(s)
The half-reactions involved are:
Reduction: Zn2+ + 2e– → Zn (E0 = –0.76 V)
Oxidation: 2Al → 2Al3+ + 6e– (E0 = –1.66 V)
To determine whether a reaction is spontaneous, we can use the following rule: A reaction is spontaneous if the overall cell potential (ΔEcell) is positive.
To find the overall cell potential (ΔEcell), we can use the equation:
ΔEcell = E0(reduction) + E0(oxidation)
a. For reaction a:
ΔEcell = E0(reduction) + E0(oxidation)
ΔEcell = (–2.37 V) + (+0.80 V) = –1.57 V
Since ΔEcell is negative (-1.57 V), the reaction a is not spontaneous as written.
b. For reaction b:
ΔEcell = E0(reduction) + E0(oxidation)
ΔEcell = (–0.76 V) + (–1.66 V) = –2.42 V
Since ΔEcell is negative (-2.42 V), the reaction b is not spontaneous as written.
Therefore, both reactions are not spontaneous as written based on the given reduction potentials.