Asked by Narasaq
The red-ox reaction is MgSO4+2K yields K2 SO4+Mg. I thought the oxidation reaction is 2 k yields K2(subscript) + e^-2. I said the reduction half reaction was Mg+e- yields Mg. Are these correct? Thanks for checking.
Answers
Answered by
DrBob222
close but no cigar.
K ==> K^+ + e or if you want to write it double then
2K ==> 2K^+ + 2electrons. Note it is 2e and not e^-2 (e^-2 denote that the electron has a charge of -2 but it has a charge of -1). You could write 2e^-1 if you wish.
Mg^+2 + 2e ==> Mg is the reduction reaction. Note also that I usually write redox and don't use a hyphen but I don't know what you've been taught by your teacher. Follow your teacher's instructions on this.
K ==> K^+ + e or if you want to write it double then
2K ==> 2K^+ + 2electrons. Note it is 2e and not e^-2 (e^-2 denote that the electron has a charge of -2 but it has a charge of -1). You could write 2e^-1 if you wish.
Mg^+2 + 2e ==> Mg is the reduction reaction. Note also that I usually write redox and don't use a hyphen but I don't know what you've been taught by your teacher. Follow your teacher's instructions on this.
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