Write a balanced half reaction for TeO3^2- (aq) = Te(s)
To write a balanced half-reaction for TeO3^2- (aq) = Te(s), we need to determine the oxidation state of each element involved.
In TeO3^2-, the oxidation state of Te is +6, and the oxidation state of O (in this case, O from TeO3^2-) is -2.
To balance the half-reaction, we need to equalize the number of electrons gained and lost. Since the oxidation state of Te is reduced from +6 to 0, it gains 6 electrons.
Therefore, the balanced half-reaction is:
TeO3^2- (aq) + 6e^- = Te(s)