Question

3. Balance the following redox equations by the ion-electron half-reaction method: (a) (acid solution): Cr0(aq) + U" (aq) → Cr" (aq) +UO; (aq) (b) (acid solution): Cr(s) + O2(g) → Cr* (aq) (C) (basic solution): P. (s) + OH' (aq) → PH; (g) + H2PO2 (aq)

Answer 1

oxidation croy - (aq) + gut (og) Cy3t + opt Reduction C207 2 er+3 x=+3 22-14= -2 22= 12 276 Decrease in ON Rocution 2 044 + 2=ty 00.2+ 2-4=+2 x=+6 in Oxidation state Oxidation Increase 3 + Reduction hay -cell Rxy 1) Balance atom other than of H crop - 27 2 Balance o/H (Acicure) add Ht and balance e bé + coo7 +1447 2683+ + 7H₂0 - Oxidation beeld cell Rxn gut + 2420 00 + 4H+ +ge ②

Multiply eg. ♡ by ③ 3oht + 6H₂O Addeg. ① and ② 300 2+ + 1247 +be ② be +(22072- +144+ 212 3+ + 74₂0 3047 + 6H₂0 - 300 ²4 +12H+ + C2072- + 304+ + 2H+ 2g 3 + + 300 2+ + H₂O this

Reduction Page No. (e) Py + oH PH2 + H₂PO2- on cotton 2 +2-4=u 2-2=-1 =-1+2 =+) OX" half cell Reaction Py - H₂PO2 Balance P atoms 4 H₂ PO, Puls Balance O-N by adding e Py - 4H2PO2 + 4e - Balance change by adding SOHO - Pu+BOHO 4H₂P07 - the O Reduction Hald cell am Balance P atoms Py 4PH₂. Balame o N by adeling e 128+ P4 - Putl2e- 4PH2 4PH2

Balonce charge by adding 12040 4 PH₂ + 1204 Put 12 e Py + 12 et 12 H₂O Multiply eq. by @ 4PH₂ + 1204 2 3Py + 24 OH - 12 H₂ Por + lze - Add e 6 and ② 3 Pu + 2404 - P4 +126 + 12 H2O - 12 H₂POj tize 4PH₂ +/2018 upu + 1204 + 12 H₂O +74 PH₂ + 12 H₂ PO₂ " thus

© ) +0318) –> (934 (ag). oxiclation * oxidation hat ull Reaction é + 4H+ + co(s) to Corst (aq) + 2H₂O since, in this ques, charge is balanced problem in this question, so the mention here but there is some most probable solution is