(a) Cl2(g) is reduced to Cl−(aq) and Sn(s) is oxidized to Sn2+(aq).
(b) Sn2+(aq) is displaced from solution by Zn(s).
(c) The cell reaction is 2Cu+(aq) --> Cu2+(aq) + Cu(s)
(d) MgI2(aq) is produced from Mg(s) and I2(s).
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Verified answer
(a)
standard electrode potentials you need to know are:
Cl₂(g) + 2e⁻ ⇄ 2Cl⁻(aq) , E° = + 1.36V
Sn²⁺(aq) + 2e⁻ ⇄ Sn(s) , E° = - 0.14V
half reactions with potentials
reduction
Cl₂(g) + 2e⁻ ⇄ 2Cl⁻(aq) , E° = + 1.36V
oxidation
Sn(s) ⇄ Sn²⁺(aq) + 2 e⁻ , E° = + 0.14V
so overall reaction
Cl₂(g) + Sn(s) ⇄ 2 Cl⁻(aq) + Sn²⁺(aq) ,
has cell potential
E° = + 1.36V + 0.14V = + 1.50V
(b)
standard electrode potentials you need to know are:
Sn²⁺(aq) + 2e⁻ ⇄ Sn(s) , E° = - 0.14V
Zn²⁺(aq) + 2e⁻ ⇄ Zn(s) , E° = - 0.76V
half reactions with potentials
reduction
Sn²⁺(aq) + 2e⁻ ⇄ Sn(s) , E° = - 0.14V
oxidation
Zn(s) ⇄ Zn²⁺(aq) + 2e⁻ , E° = + 0.76V
so overall reaction
Sn²⁺(aq) + Zn(s) ⇄ Sn(s) + Zn²⁺(aq) ,
has cell potential
E° = - 0.14V + 0.76V = + 0.62V
(c)
standard electrode potentials you need to know are:
Cu²⁺(aq) + e⁻ ⇄ Cu⁺(aq) , E° = + 0.16V
Cu⁺(aq) + e⁻ ⇄ Cu(s) , E° = + 0.52V
half reactions with potentials
reduction
Cu⁺(aq) + e⁻ ⇄ Cu(s) , E° = + 0.52V
oxidation
Cu⁺(aq) ⇄ Cu²⁺(aq) + e⁻, E° = - 0.16V
so overall reaction
2Cu⁺(aq) ⇄ Cu(s) + Cu²⁺(aq) ,
has cell potential
E° = + 0.52V - 0.16V = + 0.36V
(d)
standard electrode potentials you need to know are:
Mg²⁺(aq) + 2e⁻ ⇄ Mg(s) , E° = - 2.36V
I₂(s) + 2e⁻ ⇄ 2I⁻(aq) , E° = + 0.54V
half reactions with potentials
reduction
I₂(s) + 2e⁻ ⇄ 2I⁻(aq) , E° = + 0.54V
oxidation
Mg(s) ⇄ Mg²⁺(aq) + 2e⁻ , E° = + 2.36V
so overall reaction
I₂(s) + Mg(s) ⇄ 2I⁻(aq) + Mg²⁺(aq)
which is the same as
I₂(s) + Mg(s) ⇄ MgI₂(aq)
has cell potential
E° = + 0.54V + 2.36V = + 2.90V
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