Potassium permanganate, KMnO_4, is a powerful oxidizing agent. The products of a

Question

Potassium permanganate, KMnO_4, is a powerful oxidizing agent. The products of a given redox reaction with the permanganate ion depend on the reaction conditions used. In basic solution, the following equation represents the reaction of this ion with a solution containing sodium iodide: MnO_4^- (aq) + I^- (aq) rightarrowMnO_2 (s) + I_2 (aq) Since this reaction takes place in basic solution, H_2 O(1) and OH^- (aq) will be shown in the reaction. Places for these species are indicated by the blanks in the following restatement of the equation: MnO_4^- (aq) + I^- (aq) + rightarrowMnO_2 (s) + I_2 (aq) + What are the coefficients of the reactants and products in the balanced equation above? Remember to include H_2 O(1) and OH^- (aq) in the blanks where appropriate. Your answer should have six terms. Enter the equation coefficients in order separated by commas (e.g., 2,2,1,4,4,3). Include coefficients of 1, as required, for grading purposes. Metal plating is done by passing current through a metal solution. For example, an item can become gold plated by attaching the item to a power source and submerging it into a Au^3+ solution. The item itself serves as the cathode, at which the Au^3+ ions are reduced to Au(s). A piece of solid gold is used as the anode and is also connected to the power source, thus completing the circuit. What mass of gold is produced when 18.1 A of current are passed through a gold solution for 44.0 min ? Express your answer with the appropriate units.

Explanation / Answer

The balance reaction is as follows:
2 MnO4- + 6 I- + 4 H2O = 2 MnO2 + 3 I2 + 8 OH-

For the reaction 2Co3+ +2Cl- = 2Co2+ Cl2 E 0= 0.483 V

E = E^o - 0.0592/n log Q

Q =[ product ][reactants]

= (0.569 )^2*(6.40)/(0358)^2(0.755)^2

=28.36

n =2 since it is two electron change

E= +0.483 - 0.059/ 2 * log 28.36

E= +0.483 - 0.059/ 2 *1.453

E= 0.483-0.0429

E=+0.4401 V

Au3+(aq ) + 3e- = Au (s)

Here time = 44.0 min= 2640 s

18.1 A or 18.1 coulomb / s * 2640 s = 47784 C

One Faraday = 96,500 Coulombs
47784 Coulombs / 96,500 Coulombs / Faraday

= 0.495 Faraday

For every 3 Faradays of electricity used up , one mole Au forms.
0.495 Faraday x 1 mole Au/ 3 Faradays = 0.165 mole Au

Molar mass of Au= 196.967 g

Amount of Au = 196.967 g/ mol* 0.165 mol

=32.51 g

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