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Oxidant half reaction: MnO4- + 8 H+ + 5e- --> Mn2+ + 4 H2O
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The following reaction takes place in an acidic solution. MnO4–(aq) + Cl–(aq) Mn2+ + Cl2(g) (unbalanced) Write the reduction and oxidation half-reactions (without electrons). :(?
Reduction Half-Reaction: MnO4-(aq) → Mn2+(aq) Oxidation Half-Reaction: Cl-(aq) → Cl2(g)
Why titration of potassium permanganate is carried out in acidic medium?
This is because in acidic medium kmn04 decomposes to give Mn2+ ions which impart pink colour to the solution. The reaction being Mno4- +8H+5e ----->Mn2+ + 4H2O
Why KMnO4 titration's are carried out only in the presence of dilute H2SO4?
The permanganate ion, MnO4- purple in colour. The sulfuric acid is added to create acidic conditions. In acidic conditions, the MnO4- gets reduced to Mn2+ which is colourless. Therefore, judging by the colour change, we can figure out how much of the potassium permanganate has been used.
What is the formula of manganese III chromate?
Chromium (III) has a 3+ charge. Permanganate (MnO4-) Has a 1- charge. Therefore it takes 3 Permanganate anions to equal the charge of a Chromium (III) cation. This should be Cr(MnO4)3
What is the chemical name of Mn2 plus?
The compound Mn2(SO3)3 is called manganese(III)sulphite.
Overall redox equation fo MnO4 reacting with Fe2 in acid solution give Mn2 and Fe3?
8H+ + MnO4- + 5Fe2+--> 5Fe3+ + Mn2+ + 4H20
The following reaction takes place in an acidic solution. MnO4–(aq) + Cl–(aq) Mn2+ + Cl2(g) (unbalanced) Write the reduction and oxidation half-reactions (without electrons). :(?
Reduction Half-Reaction: MnO4-(aq) → Mn2+(aq) Oxidation Half-Reaction: Cl-(aq) → Cl2(g)
What is the SPDF Notation for Mn2 plus?
Mn: 1s22s22p63s23p63d54s2 Mn2+: 1s22s22p63s23p63d5
What is the dissociation reaction for MnO 4?
In an acidic solution: 8 H+ + MnO4− + 5 e− → Mn2+ + 4 H2O In a neutral solution: 2 H2O + MnO4− + 3 e− → MnO2 + 4 OH−
Why titration of potassium permanganate is carried out in acidic medium?
This is because in acidic medium kmn04 decomposes to give Mn2+ ions which impart pink colour to the solution. The reaction being Mno4- +8H+5e ----->Mn2+ + 4H2O
In the reaction Mg(s) plus 2H2O(I) and 61664 Mg(OH)2(s) plus H2(g) what is the oxidation state of each H in H2(g)?
Simply remember OIL RIG. Oxidation Is Loss (of electron) and Reduction Is Gain (of electron). In the case of MnO4-(aq) + 8H+(aq) + 5e- --> Mn2+(aq) + 4H2O(l) H is the oxidizing agent, because it causes Mn to be oxidized to Mn2+.
Which balanced equation represents an oxidation-reduction reaction?
Equations that separate the oxidation from the reduction parts of the reaction
What is the colour of Mn2 plus ion in estimation of iron 2 by permanganometry?
The color of Mn2+ ion in estimation of iron 2 permanganometry is purple.
Why KMnO4 titration's are carried out only in the presence of dilute H2SO4?
The permanganate ion, MnO4- purple in colour. The sulfuric acid is added to create acidic conditions. In acidic conditions, the MnO4- gets reduced to Mn2+ which is colourless. Therefore, judging by the colour change, we can figure out how much of the potassium permanganate has been used.
How many protons and electrons are in Mn2 plus?
Mn has 25 protons.
Why is titration between potassium permanganate and sodium oxalate conducted slowly?
It is a reaction between negatively charged ions so they repel each other. For the reaction to work it is first heated, the Mno4- reacts with the Mn2+ produced to form Mn3+. The Mn3+ then reacts with the ethanedoate to form co2 and Mn2+. This is an example of autocatalysis.
What is the chemical equation for the reaction between methanoic acid and acidified potassium permanganate?
The half equation for the reduction of carbon dioxide to methanoic acid and its standard electrode potential value is given as: CO2 + 2H+ + 2e- gives HCOOH EO = -0.11 V The half equation for the reduction of ACIDIFIED potassium manganate(VII) and its standard electrode potential value is given as: MnO4- + 8H+ + 5e- gives Mn2+ + 4H2O EO = +1.51 V Thus, the standard electrode potential E for the overall reaction is: E = 1.51 - (-0.11) = +1.62 V Since it is positive (and well positive), the oxidation of methanoic acid by potassium manganate(VII) can take place under standard conditions. The overall equation (obtained by balancing the number of electrons) is: 2MnO4- + 6H+ + 5HCOOH gives 2Mn2+ + 8H2O + 5CO2
