Answer:
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
