18. Reverse the reaction and divide by 2,
Eocell = (-2.10V)/2 = -1.05V ,
so option b) is the answer
19. 2 Al+3 + 6e- -------> 2Al ;
Eo = 2*(-1.66) = -3.32V
6 Cl- ---------> 3Cl2 + 6e- ;
Eo = 3*(-1.36) = -4.08V
Add these 2 above equations to get the overall 9eqn
given in the question)
Eocell = Ecathode -
Eanode
ΔGo=-nFEocell , use these formulas
to get the answer
20) At room temperature,
Ecell = Eocell -
0.0592/n log [Ox]/[Red]
Ecell = 0.460 - 0.0592/2 log (0.017 /
0.18)
Ecell = 0.4903 V option d)
21) electrons deposited = it /F = (15*160*60 sec) / 96485 C/mol =
1.49245 mole of electrons
1.49245 * (1 mole of Cr / 3moles of elcrons) = 0.49748
mol of Cr
mass of Cr deposited = 0.49748 mol of Cr * 51.99
gm/mol = 25.86 gms = 25.89 gms approx
If the value of E_cell^0 is 2.10 V for the reaction F_2 (g) + 2Fe^2+ (aq)...
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