Here, potential change depends on diffusibility of ion currents in the cell. Diffusion is a function of distance , more distance will cause slow diffusion therefore , a small cell will will have less diffusion of ion currents so a large potential change will occur than a large cell. A large due to large surface area and volume cause the ionic current to inhibit it's effect as it move farther from site of ion currents generation. Ions will diffuse as they travel long distance so there will be a small potential change.
A given intensity of injected current will produced a larger potential change in a small cell...
A typical cell has an electric potential difference across its cell membrane, The electric potential interior to the cell is 70mV less than that on the exterior. Under certain circumstances, the cell can redistribute charge so that the electric potential inside is 40 mV greater than that outside. Assuming the membrane is 12 nm thick and that the net electric field inside it is uniform, how does that field change in the transition from having an interior that is 70...
A cell has a resting membrane potential of -70mV. In each of the following cases, would the cell depolarize, or hyperpolarize? Justify your answers with a written explanation that relates to the GHK equation; you do not need to calculate new membrane potentials, but you should explain why the changes would occur. Hint: Use the simulations from class to solve these Cell becomes more permeable to Cl- Cell becomes more permeable to Ca++ (Hint: which ion shown in the simulation...
Explain why small uncertainties in the measured cell potential can cause large error in the response of ISEs.
Based on the given below figure, write down the following three
equations
(1) PV cell current equation (Apply KCL)
(2) PV Cell voltage equation (Apply KVL)
(3) Diode Current Equation (From Diode Characteristics; this
equation will be vital to simulate I-V characteristics variation
with respect to temperature change.)
.Based on the given below figure, write down the following three equations (1) PV cell current equation (Apply KCL) (2) PV Cell voltage equation (Apply KVL) (3) Diode Current Equation (From Diode...
This time you get into a snail brain neuron that is
completely quiet. The cell doesn't even begin to fire action
potentials when you inject depolarizing current, so you question
yourself if you are actually in a neuron. Nevertheless, you briefly
hyperpolarize thecell, and right after the hyperpolarization stops,
the cell fires a few action potentials that have a quite large
amplitude and then the cell becomes quite again. What could be
going on here? Your traces kind of look...
In a hydrogen fuel cell, oxygen and hydrogen are combincd to produce water and a small potential differ- ence. The steps of the chemical reaction are: at-electrode: at + electrode: H2 + 20H-→ 2H2O + 2e-; O2 + H2O + 2e-→ 2OH-, Use the table of chemical values to calculate a) the work that can be produced from this fuel cell, b) the 'waste' heat produced during standard operation, c) the voltage of the cell, and d) the minimum voltage...
A living cell extracts energy from organic "fuel" molecules in small, enzyme-catalyzed steps. This allows the process of energy extraction to be regulated. but less free energy is released from the molecules than would be released by burning them O and all of the energy is released as heat that can be used to do work in the cell O and some of the energy is stored in bonds of activated carriers O and more free energy is released from...
Calculate the cell potential, the equilibrium constant, and the free-energy change for: Ca(s)+Mn2+(aq)(1M)⇌Ca2+(aq)(1M)+Mn(s) given the following Eo values: Ca2+(aq)+2e−→Ca(s) Eo = -1.59 V Mn2+(aq)+2e−→Mn(s) Eo = -0.54 V 1.) Calculate the equilibrium constant. 2.) Free-energy change?
Calculate the cell potential, the equilibrium constant, and the free-energy change for: Ca(s)+Mn2+(aq)(1M)⇌Ca2+(aq)(1M)+Mn(s) given the following Eo values: Ca2+(aq)+2e−→Ca(s) Eo = -2.02 V Mn2+(aq)+2e−→Mn(s) Eo = -1.41 V
*Electrochemistry
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The following two reactions have ξ values given below 4°-2.07 V If the potential for the reaction pt2t following half reactions. 2e Pt is assigned a value of zero, calculate the cell potentials for the a. Ag+e-Ag -0.40V 0,4 ov 0+0.40 # b. F2cF 5.oV 4. Consider the galvanic cell at 25°C Zn Zn2 I Cu2+ Cu Calculate the ratio of IZn2+1 to [Cu2+1 when the cell potential has dropped...