Balance the following equation in acidic conditions. Phases are optional.?

2 Answers

• [6H^+] +[ 6 Cr^2+] +[H2MoO4]•••> [6Cr^3+] + Mo + 4 H2o

• so we add H+ to the left and get H2O on the right

  Cr^2+ + H2MoO4 + H+ → Cr^3+ + Mo + H2O

  Cr goes from +2 to +3 or loses one e-

  Mo is +6 in H2MoO4 so it goes to 0 gaining 6 e-

  so we must multiply Cr by 6 to balance e-

  6Cr^2+ + H2MoO4 + H+ → 6Cr^3+ + Mo + H2O

  now we balance the spectators

  take O first – 4 on left so we change the right side

  6Cr^2+ + H2MoO4 + H+ → 6Cr^3+ + Mo + 4H2O

  now we have 8 H’s on the right

  we don’t want to mess with the H2MoO4

  so we make H+ up to 6 for a total of 8 on the left

  6Cr^2+ + H2MoO4 + 6H+ → 6Cr^3+ + Mo + 4H2O

  now it’s balanced