Asked by Monique
Balance the following redox reaction occurring in acidic solution.
H+ + Cr --> H2 + Cr3+
H+ --> H2 do I add H+ to this side since H2 = 0? Also what about electrons.
Cr -> Cr3+ + 3e-
Hope I make sense
H+ + Cr --> H2 + Cr3+
H+ --> H2 do I add H+ to this side since H2 = 0? Also what about electrons.
Cr -> Cr3+ + 3e-
Hope I make sense
Answers
Answered by
DrBob222
H^+ ==> H2
First balance the H atoms so we are looking at the same number on each side. That's about the only way to balance redox equations.You do that so you can talk about total charges on each side.
2H^+ ==> H2
H has changed from +2(+1 each) to zero; therefore add 2e to the left.
2H^+ + 2e ==> H2
Cr is correct at Cr==> Cr^3+ + 3e.
Then you multiply the H equation by 3 and the Cr equation by 2 and add them. Cancel the electrons (there will be 6 on each side) and you have it.
First balance the H atoms so we are looking at the same number on each side. That's about the only way to balance redox equations.You do that so you can talk about total charges on each side.
2H^+ ==> H2
H has changed from +2(+1 each) to zero; therefore add 2e to the left.
2H^+ + 2e ==> H2
Cr is correct at Cr==> Cr^3+ + 3e.
Then you multiply the H equation by 3 and the Cr equation by 2 and add them. Cancel the electrons (there will be 6 on each side) and you have it.
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