I is +5 in IO3^-
B is correct.
What's the problem with C?
Al^3+ + 3e ==> Al is one half cell
The other one is
2IO3^- + 10 e + 12H^+ = I2 +6H2O
I assume an acid solution.
CONSIDER THE FOLLLOWING UNBALANCED REDOX EQUATION:
IO3^-(AQ) + Al(S)-->I2(S) + A;^3+(AQ)
A. WHAT IS THE OXDATION NUMBER OF IODINE IN THE IODATE ION? I GOT IO3^6(AQ)?
B. WHICH SPECIES IS THE REDUCING AGENT I GOT AL^3+(AQ)
C. USING THE OXIDATION AND REDUCTION HALF REACTIONS WRITE THE BALANCED REDOX EQUATION
1 answer
1 answer