Draw out a Simple electro Chemical Symbols in ions​

In redox reactions, electrons are transferred from one species to another. If the reaction is spontaneous, energy is released, which can then be used to do useful work. To harness this energy, the reaction must be split into two separate half reactions: the oxidation and reduction reactions. The reactions are put into two different containers and a wire is used to drive the electrons from one side to the other. In doing so, a Voltaic/ Galvanic Cell is created.

Explanation:

When a redox reaction takes place, electrons are transferred from one species to the other. If the reaction is spontaneous, energy is released, which can be used to do work. Consider the reaction of a solid copper (Cu(s)) in a silver nitrate solution (AgNO3(s)).

he AgNO3(s) dissociates in water to produce Ag+(aq) ions and NO­­−3(aq) ions. The NO3-(aq) ions can be ignored since they are spectator ions and do not participate in the reaction. In this reaction, a copper electrode is placed into a solution containing silver ions. The Ag+(aq) will readily oxidize Cu(s) resulting in Cu2+(aq), while reducing itself to Ag(s).

This reaction releases energy. When the copper electrode solid is placed directly into a silver nitrate solution, however, the energy is lost as heat and cannot be used to do work. In order to harness this energy and use it do useful work, we must split the reaction into two separate half reactions; The oxidation and reduction reactions. A wire connects the two reactions and allows electrons to flow from one side to the other. In doing so, we have created a Voltaic/ Galvanic Cell.

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol