The quantity of electricity (in Faraday) required to completely reduce 0.2 mol Mno4- to Mn2+ ions is
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Answered by
11
Hello dear,
● Answer = 1 F
● Explaination-
# Given-
n = 0.2 mol
# Solution-
Charge on Mn in MnO4-,
q1 = -1 - 4(-2)
q1 = -1+8
q1 = +7
Charge on Mn in Mn2+,
q2 = +2
Difference in charge,
Q = q1-q2 = 7 - 2 = 5 F
But for 0.2 moles,
Electricity required will be
Q' = n × Q
Q' = 0.2 × 5
Q' = 1 F.
Electricity required in reduction reaction is 1 F.
● Answer = 1 F
● Explaination-
# Given-
n = 0.2 mol
# Solution-
Charge on Mn in MnO4-,
q1 = -1 - 4(-2)
q1 = -1+8
q1 = +7
Charge on Mn in Mn2+,
q2 = +2
Difference in charge,
Q = q1-q2 = 7 - 2 = 5 F
But for 0.2 moles,
Electricity required will be
Q' = n × Q
Q' = 0.2 × 5
Q' = 1 F.
Electricity required in reduction reaction is 1 F.
Answered by
0
Answer:0.6 F
Explanation:
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