Advertisements
Advertisements
Question
Balance the following ionic equations.
\[\ce{Cr2O^{2-}7 + Fe^{2+} + H+ -> Cr^{3+} + Fe^{3+} + H2O}\]
Advertisements
Solution
Step 1: Separate the equation into two half-reactions.
The oxidation number of various atoms are shown below:
\[\ce{\overset{+6}{Cr2}\overset{-2}{O}^{2-}7 + \overset{+2}{Fe}^{2+} + \overset{+1}{H+} -> \overset{+3}{Cr}^{3+} + \overset{+3}{Fe}^{3+} + \overset{+1}{H2}\overset{-2}{O}}\]
In this case, chromium undergo reduction, oxidation number decreases from +6 (in \[\ce{Cr2O^{2-}7}\]) to +3 (in \[\ce{Cr^{3+}}\])
\[\ce{Fe^{2+}}\] (O.N. = +2) changes to \[\ce{Fe^{3+}}\] (O.N. = +3). The species undergoing oxidation and reduction are:
Oxidation: \[\ce{Fe^{2+} -> Fe^{3+}}\]
Reduction: \[\ce{Cr2O^{2-}7 -> Cr^{3+}}\]
Step 2: Balance each half-reaction separately as:
(a) \[\ce{Fe^{2+} -> Fe^{3+}}\]
(i) Balance all atoms other than \[\ce{H}\] and \[\ce{O}\]. This step is not needed, because, it is already balanced.
(ii) The oxidation number on left is +2 and on right is +3. To account for the difference, the electron is added to the right as: \[\ce{Fe^{2+} -> Fe^{3+} + e-}\]
(iii) Charge is already balanced.
(iv) No need to add \[\ce{H}\] or \[\ce{O}\].
The balanced half-equation is:
\[\ce{Fe^{2+} -> Fe^{3+} + e-}\] ......(i)
Consider the second half-equation
(b) \[\ce{Cr2O^{2-}7 -> Cr^{3+}}\]
(i) Balance the atoms other than \[\ce{H}\] and \[\ce{O}\].
\[\ce{Cr2O^{2-}7 -> 2Cr^{3+}}\]
(ii) The oxidation number of chromium on the left is +6 and on the right is +3. Each chromium atom must gain three electrons. Since there ar two \[\ce{Cr}\] atoms, add \[\ce{6e-}\] on the left.
\[\ce{Cr2O^{2-}7 + 6e^{-} -> 2Cr^{3+}}\]
(iii) Since the reaction takes place in acidic medium add \[\ce{14H+}\] on the left to equate the net charge on both sides.
\[\ce{Cr2O^{2-}7 + 6e^{-} + 14H^{+} -> 2Cr^{3+}}\]
(iv) To balance \[\ce{FI}\] atoms, add \[\ce{7H2O}\] molecu;es on the right.
\[\ce{Cr2O^{2-}7 + 6e^{-} + 14H^{+} -> 2Cr^{3+} + 7H2O}\] ......(ii)
This is the balanced half-equation.
Step 3: Now add up the two half-equations. Multiply equation (i) by 6 so that electrons are balanced.
\[\ce{Fe^{2+} -> Fe^{3+} + e- × 6}\]
\[\ce{Cr2O^{2-}7 + 6e- + 14H+ -> 2Cr^{3+} + 7H2O}\]
\[\ce{6Fe^{2+} + Cr2O^{2-}7 + 14H+ -> 6Fe^{3+} + 2Cr^{3+} + 7H2O}\]
The balanced equation is: \[\ce{6Fe^{2+} + Cr2O^{2-}7 + 14H+ -> 6Fe^{3+} + 2Cr^{3+} + 7H2O}\]
APPEARS IN
RELATED QUESTIONS
Calculate the oxidation number of sulphur, chromium and nitrogen in H2SO5, `"Cr"_2"O"_7^(2-)` and `"NO"_3^-`. Suggest structure of these compounds. Count for the fallacy.
In Ostwald’s process for the manufacture of nitric acid, the first step involves the oxidation of ammonia gas by oxygen gas to give nitric oxide gas and steam. What is the maximum weight of nitric oxide that can be obtained starting only with 10.00 g. of ammonia and 20.00 g of oxygen?
Justify that the following reaction is redox reaction; identify the species oxidized/reduced, which acts as an oxidant and which acts as a reductant.
\[\ce{2Cu2O_{(S)} + Cu2S_{(S)}->6Cu_{(S)} + SO2_{(g)}}\]
Balance the following reaction by oxidation number method.
\[\ce{H2SO4_{(aq)} + C_{(s)} -> CO2_{(g)} + SO2_{(g)} + H2O_{(l)}(acidic)}\]
Balance the following reaction by oxidation number method.
\[\ce{Bi(OH)_{3(s)} + Sn(OH)^-_{3(aq)}->Bi_{(s)} + Sn(OH)^2-_{6(aq)}(basic)}\]
Balance the following redox equation by half-reaction method.
\[\ce{H2C2O_{4(aq)} + MnO^-_{4(aq)}->CO2_{(g)} + Mn^2+_{( aq)}(acidic)}\]
Balance the following redox equation by half-reaction method.
\[\ce{Bi(OH)_{3(s)} + SnO^2-_{2(aq)}->SnO^2-_{3(aq)} + Bi^_{(s)}(basic)}\]
Identify the oxidising agent in the following reaction:
\[\ce{CH4_{(g)} + 2O2_{(g)} -> CO2_{(g)} + 2H2O_{(l)}}\]
When methane is burnt completely, oxidation state of carbon changes from ______.
Write balanced chemical equation for the following reactions:
Permanganate ion \[\ce{(MnO^{-}4)}\] reacts with sulphur dioxide gas in acidic medium to produce \[\ce{Mn^{2+}}\] and hydrogen sulphate ion.
Write balanced chemical equation for the following reactions:
Reaction of liquid hydrazine \[\ce{(N2H4)}\] with chlorate ion \[\ce{(ClO^{-}3)}\] in basic medium produces nitric oxide gas and chloride ion in gaseous state.
Balance the following equations by the oxidation number method.
\[\ce{I2 + NO^{-}3 -> NO2 + IO^{-}3}\]
Balance the following equations by the oxidation number method.
\[\ce{MnO2 + C2O^{2-}4 -> Mn^{2+} + CO2}\]
Balance the following ionic equations.
\[\ce{MnO^{-}4 + SO^{2-}3 + H^{+} -> Mn^{2+} + SO^{2-}4 + H2O}\]
Balance the following ionic equations.
\[\ce{MnO^{-}4 + H^{+} + Br^{-} -> Mn^{2+} + Br2 + H2O}\]
Consider the following reaction:
\[\ce{xMnO^-_4 + yC2O^{2-}_4 + zH^+ -> xMn^{2+} + 2{y}CO2 + z/2H2O}\]
The values of x, y, and z in the reaction are, respectively:
