standard cell potential and is it spontaneous under standard state conditions?

Question

Calculate the standard cell potential for each reaction below, and note whether the reaction is spontaneous under standard state conditions.

(a) Mn(𝑠)+Ni2+(𝑎𝑞)⟶Mn2+(𝑎𝑞)+Ni(𝑠)Mn(s)+Ni2+(aq)⟶Mn2+(aq)+Ni(s)

(b) 3Cu2+(𝑎𝑞)+2Al(𝑠)⟶2Al3+(𝑎𝑞)+3Cu(𝑠)3Cu2+(aq)+2Al(s)⟶2Al3+(aq)+3Cu(s)

(c) Na(𝑠)+LiNO3(𝑎𝑞)⟶NaNO3(𝑎𝑞)+Li(𝑠)Na(s)+LiNO3(aq)⟶NaNO3(aq)+Li(s)

(d) Ca(NO3)2(𝑎𝑞)+Ba(𝑠)⟶Ba(NO3)2(𝑎𝑞)+Ca(𝑠)Ca(NO3)2(aq)+Ba(s)⟶Ba(NO3)2(aq)+Ca(s)

Answer 1

(a) Mn(𝑠) + Ni²⁺(𝑎𝑞) ⟶ Mn²⁺(aq) + Ni(s)

Standard reduction potentials:

Mn²⁺ ==> Mn Eº = -1.185 V

Ni²⁺ ==> Ni Eº = -0.257 V

Ni2+ will be reduced at the cathode and Mn will be oxidized at the anode

Eºcell = cathode - anode = -0.257 - (-1.185) = 0.928 V (spontaneous)

(b) 3Cu²⁺(𝑎𝑞) + 2Al(𝑠)==> 2Al³⁺(aq) + 3Cu(s)

Cu²⁺ ==> Cu Eº = +0.337 V (cathode)

Al³⁺ ===> Al Eº = -1.662 V (anode)

Eºcell = cathode - anode = 0.337 - (-1.662) = 1.325 V (spontaneous)

(c) Na⁺ ==> Na Eº = -2.71 V (cathode)

Li⁺ ==> Li Eº = -3.04 V (anode)

Eºcell = cathode - anode = -2.71 - (-3.04) = 0.33 (spontaneous)

(d) Ca²⁺ ==> Ca Eº = -2.87 V(cathode)

Ba²⁺ ==> Ba Eº = -2.91 V (anode)

Eºcell = cathode - anode = -2.87 - (-2.91) = 0.04 V (spontaneous)