Question Solved1 Answer (1) Identify each of the following half-reactions as either an oxidation half-reaction or a reduction half-reaction. half-reaction identification I2(s) + 2e21(aq) Cr(s)Cr**(aq) + 3e" (2) Write a balanced equation for the overall redox reaction. Use smallest possible integer coefficients. ,(1) Identify each of the following half-reactions as either an oxidation half-reaction or a reduction half-reaction. half-reaction identification Hg2*(aq) + 2e–Hg(1) Cu(s)- →Cu2*(aq) + 2e¯ (2) Write a balanced equation for the overall redox reaction. Use smallest possible integer coefficients.

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Transcribed Image Text: (1) Identify each of the following half-reactions as either an oxidation half-reaction or a reduction half-reaction. half-reaction identification I2(s) + 2e21(aq) Cr(s)Cr**(aq) + 3e" (2) Write a balanced equation for the overall redox reaction. Use smallest possible integer coefficients. ,(1) Identify each of the following half-reactions as either an oxidation half-reaction or a reduction half-reaction. half-reaction identification Hg2*(aq) + 2e–Hg(1) Cu(s)- →Cu2*(aq) + 2e¯ (2) Write a balanced equation for the overall redox reaction. Use smallest possible integer coefficients.
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Transcribed Image Text: (1) Identify each of the following half-reactions as either an oxidation half-reaction or a reduction half-reaction. half-reaction identification I2(s) + 2e21(aq) Cr(s)Cr**(aq) + 3e" (2) Write a balanced equation for the overall redox reaction. Use smallest possible integer coefficients. ,(1) Identify each of the following half-reactions as either an oxidation half-reaction or a reduction half-reaction. half-reaction identification Hg2*(aq) + 2e–Hg(1) Cu(s)- →Cu2*(aq) + 2e¯ (2) Write a balanced equation for the overall redox reaction. Use smallest possible integer coefficients.
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Step 1Since you have posted multiple question , as per guidelines we can answer one per session . If you want remaining please repost.In this session , I have solved :Step 2Calculations:I2(s) + 2e----> 2I-(aq) Since iodine is gaining electrons and gain of electrons is termed as reduction .Hence , this is reduction half cell.Multiply this equation with 3.3I2(s) + 6e----> 6I-(aq) ---1C ... See the full answer