Question

# 1- 30.00 mL of sodium sulfite solution are titrated by 45.78 mL of 0.000983 M potassium...

1- 30.00 mL of sodium sulfite solution are titrated by 45.78 mL of 0.000983 M potassium dichromate in the presence of sulfuric acid, generating sulfate and chromium (III) ions. Determine the balanced net ionic equation for the titration, and calculate the molarity of the sodium sulfite solution.

2- A voltaic cell is constructed at 25˚C using half cell couples of Fe3+/Fe2+ with 1.00M/0.100 M concentrations and MnO4 - /Mn2+ at 1.00x10-2 M/1.00x10-4 M concentrations in an acidic solution of 1x10-3 M hydronium. Find the electrode potential for each half cell, and the overall cell potential.

1.

Net ionic equation between dichromate and sulphite solutions is,

Cr2O72- (aq.) + 3 SO32- (aq.) + 8 H+ (aq.) 2 Cr3+ (aq.) + 3 SO42- (aq.) + 4 H2O (l)

Useful formula is,

M1 V1 / n1 = M2 V2 / n2

Given data,

Molarity of idchromate, M1 = 0.000983 M

Volume of dichromate, V1 = 45.78 mL

Moles of dichromate, n1 = 1 mol

Molarity of sulphite solution, M2 = ?

Volume of sulphite solution, V2 = 30.00 mL

Moles of sulphite, n2 = 3 mol

Therefore,

0.000983 * 45.78 / 1 = M2 * 30.00 / 3

M2 = Molarity of sulphite solution = 0.00450 M

#### Earn Coins

Coins can be redeemed for fabulous gifts.