1896 United States presidential election in Kansas

1896 United States presidential election in Kansas

← 1892 November 3, 1896 1900 →
 
Nominee William Jennings Bryan William McKinley
Party Democratic Republican
Alliance Populist
Home state Nebraska Ohio
Running mate Arthur Sewall Garret Hobart
Electoral vote 10 0
Popular vote 171,675 159,345
Percentage 51.32% 47.63%

County Results

President before election

Grover Cleveland
Democratic

Elected President

William McKinley
Republican

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

Kansas was won by the Democratic nominees, former U.S. Representative William Jennings Bryan of Nebraska and his running mate Arthur Sewall of Maine. They Republican nominees, former Governor of Ohio William McKinley and his running mate Garret Hobart of New Jersey. Bryan won the state by a narrow margin of 3.69%.

With his win in the state, Bryan became the first Democratic presidential candidate to win the state of Kansas. Bryan would later lose Kansas to McKinley four years later during their rematch and would later lose the state again to William Howard Taft in 1908. This remains the only election since Kansas statehood in which the Republican candidate won the presidency without Kansas, or where Kansas voted more Democratic than Maryland. This was the last election until 2020 in which a Democrat carried Johnson County with a majority.

This was the first of six elections in which the Democratic nominee carried Kansas, traditionally a Republican bastion like the other Plains States. The others were 1912, 1916, 1932, 1936, and 1964.[1]

  1. ^ "Historical U.S. Presidential Elections 1789-2020 - 270toWin". 270toWin.com. Retrieved April 14, 2022.

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