ate the balanced half-equation for the reduction of dichromate ions to chromium(III) ions.
Question
ate the balanced half-equation for the reduction of dichromate ions to chromium(III) ions.
Solution
The balanced half-equation for the reduction of dichromate ions (Cr2O7^2-) to chromium(III) ions (Cr^3+) in acidic solution is:
Cr2O7^2- + 14H^+ + 6e^- → 2Cr^3+ + 7H2O
Here's the step-by-step process:
-
Write the unbalanced equation for the reaction: Cr2O7^2- → Cr^3+
-
Balance the chromium atoms: Cr2O7^2- → 2Cr^3+
-
Balance the oxygen atoms by adding water molecules to the side that needs oxygen: Cr2O7^2- → 2Cr^3+ + 7H2O
-
Balance the hydrogen atoms by adding H^+ ions to the side that needs hydrogen: Cr2O7^2- + 14H^+ → 2Cr^3+ + 7H2O
-
Balance the charge by adding electrons (e^-) to the side that is more positive: Cr2O7^2- + 14H^+ + 6e^- → 2Cr^3+ + 7H2O
This equation is balanced in terms of both mass (atoms of each element) and charge.
Similar Questions
The oxidation state of chromium in the final product formed by the reaction between KI and acidified potassium dichromate solution is+4+6+2+3
Which chromium ion is present in the following compound?Chromium (III) sulfateCr+CrCr3+Cr2+
explain about chemistry of chromine inter 2year
Acidified potassium dichromate is treated with hydrogen sulphide. In the reaction, the oxidation number of chromiumIncreases from +3 to +6Decreases from +6 to +3Remains unchangedDecreases from +6 to +2
12 A student adds 3 mol of acidified K 2Cr 2O7 to an excess of I– ions.The chromium is all reduced to Cr 3+ and I– ions are oxidised to I2.The I2 released is reduced back to I– ions by X mol of S 2O32– ions.1 mol of I2 is reduced by 2 mol of S 2O32– ions.What is the value of X?A 3 B 6 C 9 D 18
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.