Question

1) The value of E˚ for the following reaction is 1.10 V. What is the value of Ecell when the concentration of Ag+ is 0.12 M and the concentration of Cr3+ is 0.40 M?

Cr (s) + 3Ag+ (aq) à 3Ag (s) + Cr3+ (aq) E˚ = 1.54 V

2) How long will it take to plate out 2.19 g of chromium metal (52.0 g/mol) from a solution of CrBr3, using a current of 35.2 amps?

Answer 1

1)

Number of electron being transferred in balanced reaction is 3

So, n = 3

use:

E = Eo - (2.303*RT/nF) log {[Cr3+]^1/[Ag+]^3}

Here:

2.303*R*T/F

= 2.303*8.314*298.0/96500

= 0.0591

So, above expression becomes:

E = Eo - (0.0591/n) log {[Cr3+]^1/[Ag+]^3}

E = 1.1 - (0.0591/3) log (0.4^1/0.12^3)

E = 1.1-(4.66*10^-2)

E = 1.053 V

Answer: 1.05 V

2)

Electrolysis equation is:

Cr3+ + 3e- ------> Cr

1 mol of Cr requires 3 mol of electron

1 mol of electron = 96485 C

So,1 mol of Cr requires 289455 C

let us calculate mol of element deposited:

use:

number of mol, n = mass/molar mass

= 2.19/52

= 4.212*10^-2 mol

total charge = mol of element deposited * charge required for 1 mol

= 4.212*10^-2*2.895*10^5

= 1.219*10^4 C

use:

time = Q/i

= 1.219*10^4/35.2

= 3.463*10^2 seconds

= 3.463*10^2/60 min

= 5.772 min

Answer: 5.77 min