Question

Given the following standard reduction potentials: Tl(3+) + e(-) ---> Tl(2+) E°= -0.37 V Tl(3+) + 2e(-) ---> Tl(+) E°= 1.25 V

(a) Calculate the half-cell potential for the half-reaction Tl(2+) + e(-) ---> Tl(+)

(b) Determine of the following disproportionate reaction will occur sponta- neously in aqueous solution in the standard condition. 2Tl(2+)(aq) ---> Tl(3+)(aq) + Tl(+)(aq)

Answer 1

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a) Write down the equations systematically - Tl^2+ + e^- rightarrow Tl^+ equation (1) E_1 degree = ? Tl^3+ + e^- rightarrow Tl^2+ equation (2) E_2 degree = -0.37 Tl^3+ + 2e^- rightarrow Tl^+ equation (3) E_3 degree = + 1.25 Observe that on adding (1) & (2) we can get eqn (3). This implies that delta G_1 + delta G_2 = delta G_3 or delta G_1 = delta G_3 - delta G_2 And, we know delta G = -n F E degree so, -(1) FE_1 degree = -(1) F (1.25) - [- (2) F (-0.37)] -E_1 degree = -1.25 - 0.74 (Divided by F) E_1 degree = + 1.99 V b) For this, simply construct a Latina diagram put highest oxidized specials at left and move towards lowest oxidized species. Write the reduction potentials above the arrows Tl^3+ rightarrow^-0.37 Tl^2+ rightarrow^+1.99 Tl^+ Now, we need to check for Tl^2+ Key - point: