Given; 3I2(s)+5Cr2O7^2- (aq) +34H^+ (aq)------>10 Cr^3+ (aq) +6 IO3^- (aq) +17 H

ID: 699398 • Letter: G

Question

Given;

3I2(s)+5Cr2O7^2- (aq) +34H^+ (aq)------>10 Cr^3+ (aq) +6 IO3^- (aq) +17 H2O (l)

[Cr2O72-] = 0.010 M, [IO3^-]=0.0001 M, and [Cr^3+]=0.0010 M

pH = 1.0

Step 1

Find the value of Q

Q = ([ Cr3+]10 [IO3-]6 ) / [(Cr2o72-]5 [H+]34)

Find out H+ ion concentration using pH and then use all the values to get Q.

pH=-log[H+]

[H+] =10(-pH)

[H+ ]= 0.1 M

Calculation of Q

Q = ((0.0010)10 (0.0001)6 ) / (0.010)5 (0.1)34 )

Q=1.0 x10-10

Step 2

Calculate standard emf.

E0cell = Ecathode – Eanode

Oxidation takes place at anode and reduction at cathode.

Cr2O7 + 14H+ + 6e 2Cr3+ + 7H2O E0red =1.232 V

2IO3 + 12H+ + 10e I2 + 6H2O E0red = 1.195 V

E0cell = Ecathode – Eanode = 1.232 – 1.195 = 0.037 V

Step 3

Use Nernst equation to calculate emf of cell.

Ecell = E0cell - 0.0951 / n x log Q

Here n is number of electrons required in reaction = 30

Lets plug in the values,

Ecell = 0.037 V – (0.0591/30) log ( 1.0 E-10)

=0.0567 V

Calculation of delta G.

Delta G = - nEF

F is faradays constant ( 96487 C/mol)

Delta G = - 30 * 0.0567 * 96487 = -164124.4 J /mol

Since the value of delta G is negative, the reaction is spontaneous.

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