1896 United States presidential election in Arkansas

1896 United States presidential election in Arkansas
Bryan 50–60% 60–70% 70–80% 80–90% 90–100%	McKinley 50–60%
President before election; Grover Cleveland; Democratic	Elected President ; William McKinley; Republican

The 1896 United States presidential election in Arkansas took place on November 3, 1896. All contemporary 45 states were part of the 1896 United States presidential election. Voters chose eight electors to the Electoral College, which selected the president and vice president.

This election marks the establishment of one-party hegemony in every county in Arkansas except Unionist Ozark Newton and Searcy where Republicans would control local government permanently.^[1] The new secret ballot and an onerous poll tax had already severely reduced black and poor white voting at the time of this election,^[2] in the process largely eliminating opposition party challenges.

Arkansas was won by the Democratic nominees, former U.S. Representative William Jennings Bryan of Nebraska and his running mate Arthur Sewall of Maine. Three electors cast their vice presidential ballots for Thomas E. Watson. They defeated Republican nominees, former Ohio Governor William McKinley and his running mate Garret Hobart of New Jersey. Bryan won the state by a landslide margin of 48.6%.

Bryan would later defeat McKinley again in Arkansas four years later and would later win the state against William Howard Taft in 1908.

^ See Urwin, Cathy Kunzinger. Agenda for Reform: Winthrop Rockefeller as Governor of Arkansas, 1967-71. p. 32. ISBN 1557282005.
^ Perman, Michael. Struggle for Mastery: Disfranchisement in the South, 1888-1908. p. 65. ISBN 0807860255.

[1] See Urwin, Cathy Kunzinger. Agenda for Reform: Winthrop Rockefeller as Governor of Arkansas, 1967-71. p. 32. ISBN 1557282005.

[2] Perman, Michael. Struggle for Mastery: Disfranchisement in the South, 1888-1908. p. 65. ISBN 0807860255.

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