A current of 0.44 A is passed through a solution of a ruthenium nitrate salt, causing reduction of the metal ion to the metal. After 25.0 minutes, 0.345 g of Ru(s) has been deposited. A. What is the oxidation state of ruthenium in the nitrate salt? B. What is the charge on the ruthenium ion, Ru? C. What is the formula for ruthenium nitrate?

A. Oxidation state of Ru is +2

B. Positive (+)

C. Ru(NO3)2

Explanation:

First, let us calculate the quantity of electricity that will deposit 0.345g of ruthenium. This is illustrated below:

I = 0.44 A

t = 25mins = 25 x 60 = 1500secs

Q =?

Q = It

Q = 0.44 x 1500

Q = 660C

A. Molar Mass of Ru = 101g/mol

0.345g of Ru was deposited by 660C.

Therefore, 101g of Ru will be deposited by = (101 x 660)/0.345 = 193217.4C

Recall: 96500C = 1 faraday

193217.4C = 193217.4/96500 = 2 faraday.

The equation is given by

Ru^2+ 2e- —> Ru

Therefore, the oxidation state of Ru in the nitrate salt is +2

B. The charge on the ruthenium ion is positive (+)

C. The formula for the compound is given by

Ru has valency of +2

NO3 has valency of -1

During bonding, there will be an exchange of valency so that the compound formed will have a formula as Ru(NO3)2

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